Binomial Theorem Obvious is the most dangerous word in mathematics......... Bell, Eric Temple
“
”
Any algebraic expression which contains two dissimilar terms is called binomial expression. 1
For example : x + y, x 2y +
xy 2
,3
–
x 2 ! 1 +
x,
1 ( x 3 ! 1)1 / 3
etc.
or n! is pronounced as factorial n and is defined as
Factorial notation :
$n(n ' 1)(n ' 2)........ 3 . 2 .1 ; if n & N n! = # 1 ; if n % 0 " n&
Note : n! = n . (n – 1)! ;
N n r
The term nCr denotes number of combinations of r things choosen n! , n & N, r & W, 0 ) r ) n (n ' r )! r!
from n distinct things mathematically mathematically,, nCr =
/ n , Note : Other symbols of of nC r are -- ** and C(n, r). . r + n
:
r
(i)
n
Cr = nCn
Note : If nCx = nCy (ii)
n
(iii)
n
Cr
Cr '1
r
0
C r + nCr n
–
–
Either x = y
1
= n r
=
n+1
or
x+y=n
Cr
n ' r !1 r
(v)
If n and r are are rela relati tiv vely ely pri prime me,, th then nCr is divisible by n. But converse is not necessarily true.
Cr
–
1
=
n – 2
n(n ' 1)(n ' 2).........(n ' (r ' 1)) r (r ' 1)(r ' 2).......2 .1
n
Cr =
n –1
n(n ' 1) r(r ' 1)
(iv)
Cr
–
2
= ............. =
(a + b) n = nC0 anb0 + nC1 an 1 b1 + nC 2 a n 2 b2 +...+ nCr an r b r +...... + nCn a 0 bn –
–
–
where n & N n
or
(a + b)n =
1
n
C r an 'r b r
r%0
Note : If we put a = 1 and b = x in the above binomial expansion, then or (1 + x)n = nC0 + nC1 x + nC2 x 2 +... + nCr x r +...+ nCn x n n
or
( 1 + x )n =
1
n
Cr x r
r%0
JEE(MAIN) BINOMIAL THEOREM - 1
Regarding Pascal’s Triangle, we note the following : (a) (a) Each Each row of the the trian triangl gle e begi begins ns with with 1 and and ends ends with with 1. (b) (b) Any Any entry entry in a row row is is the sum sum of two two entr entrie ies s in the the prec preced eding ing row, row, one on the the immedia immediate te left left and and the other on the imm ediate right. Example # 3 : The number of dissimilar terms in the expansion of (1 – 3x + 3x 2 – x3)20 is ( A) 2 1 ( B) 31 (C) 41 (D) 61 Solution :
(1 – 3x + 3x 2 – x 3)20 = [(1 – x) 3]20 = (1 – x) 60 Therefore number of dissimilar terms in the expansion of (1 – 3x + 3x 2 – x3)20 is 61.
(x + y)n = nC0 x n y 0 + nC1 xn 1 y 1 + ...........+ nCr xn r y r + ..........+ n Cn x 0 y n (r + 1)th term is called general term and denoted by Tr+1 . T r+1 = nCr x n r yr –
–
–
Note : The rth term from the end is equal to the (n – r + 2) th term from the begining, i.e. Example # 4 : Find
Solution :
(i)
28th term of (5x + 8y)30
(i)
T27 + 1 = 30 C27 (5x)30
( i i)
/ 4 x 5 , ' * 7t h te r m of . 5 2 x +
–
T6 + 1
/ 4x , * = 9 C6 . 5 +
9 '6
27
(8y)27 =
n
Cn
/ 4x 5 , ' * 7th term of . 5 2x +
(ii)
–
r + 1
xr
–
1
yn
–
r + 1
9
30 ! (5x)3 . (8y)27 3 ! 27 !
9
/ 5 , -' * . 2x +
6
9 ! / 4x , * = 3! 6! . 5 +
3
/ 5 , - * . 2x +
6
10500
=
x3
Example # 5 : Find the number of rational terms in the expansion of (91/4 + 81/6 ) 1000. Solution :
2
The general term in the expansion of 91 / 4 1000'r
Tr+1
/ 1 , - 4* = 1000Cr - 9 * . +
! 81 / 6 3
1000
is
r
/ 1 , 1000 'r r -8 6 * 1000 2 2 2 - * = Cr 3 . +
The above term will be rational if exponent of 3 and 2 are integers 1000 ' r r and must be integers 2 2 The possible set of values of r is {0, 2, 4, ............, 1000} Hence, number of rational terms is 501
It means
th
(a)
/ n ! 2 , * term. If n is is eve even, n, ther there e is is onl only y one one midd middle le term term,, whi which ch is . 2 +
(b)
/ n ! 1 , / n ! 1 , ! 1* terms. * and If n is is odd odd,, the there re are are two two midd middle le term terms, s, which hich are are . 2 + . 2 +
th
th
Example # 6 : Find the middle term(s) in the expansion of 14
(i)
/ x 2 , -1 ' * * . 2 +
(ii)
3 , / - 3a ' a * 6 +* .
9
JEE(MAIN) BINOMIAL THEOREM - 3
When
Case -
( i)
n !1 is an integer (say m), then a 1! b
Tr+1 > T r when r < m (r = 1, 2, 3 ...., m – 1) i. e. T 2 > T1, T3 > T2, ......., Tm > Tm 1 T r+1 = T r when r = m i. e. T m+1 = Tm T r+1 < T r when r > m (r = m + 1, m + 2, .. ..... .. .....n ) i. e. T m+2 < T m+1 , T m+3 < T m+2 , ..........T n+1 < T n –
( i i) ( i i i) Conclusion :
n !1 a is an integer, say m, then T m and T m+1 will be numerically greatest terms (both terms are When 1! b equal in magnitude)
Case When
( i)
n !1 is not an integer (Let its integral part be m), then a 1! b Tr+1 > T r
i. e. ( i i)
when
r<
n !1 a 1! b
(r = 1, 2, 3,........, m –1, m)
T 2 > T 1 , T 3 > T 2, .............., Tm+1 > T m
T r+1 < T r
i. e.
when r >
n !1 a 1! b
(r = m + 1, m + 2, ..............n)
T m+2 < Tm+1 , T m+3 < Tm+2 , .............., Tn +1 < Tn
Conclusion : When
n !1 is not an integer and its integral part is m, then Tm+1 will be the numerically greatest a 1! b
term. Note : (i) (i)
In any any bin binomi omial al exp expan ansi sion on,, the the middl middle e term( term(s) s) has has grea greate test st bino binomia miall coef coeffic ficie ient nt.. n In the expansion of (a + b) If n N o . o f g rea t e s t b i n o m i al c o ef f i ci e n t G re a t e s t b i n o m i a l c o e f f i c i e n t n Even 1 Cn/2 n Odd 2 C (n 1)/2 and nC(n + 1)/2 (Values (Values of both these coefficients are equal ) (ii) (ii) In orde orderr to obta obtain in the the tterm erm havi having ng numeri numerical cally ly greates greatestt coeffi coefficie cient, nt, put a = b = 1, 1, and and procee proceed d as discussed above. 1 Example # 8 : Find the numerically greatest term in the expansion of (3 – 5x) 15 when x = . 5 Solution : Let rth and (r + 1)th be two consecutive terms in the expansion of ( 3 – 5x) 15 T r + 1 4 Tr 15 Cr 315 r (| – 5x|) r 4 15 Cr 1 315 (r 1) (| – 5x|) r 1 –
–
–
–
–
–
215)! 3. 215)! | – 5x | 4 (15 ' r ) ! r ! (16 ' r ) ! (r ' 1) ! 1 (16 – r) 5 16 – r 4 3r 4r ) 16 16
5.
4 3r r ) 4 JEE(MAIN) BINOMIAL THEOREM - 5
Self practice problems : (8)
If n is is a posi positi tive ve inte intege gerr, the then n sho show w that that 32n + 1 + 2 n + 2 is divisible by 7.
(9)
Wh ha at is the r em ai ainder wh when 7103 is divided by 25 .
(10) (10)
Find Find the last last digit, digit, last last two two digi digits ts and and last last three three digi digits ts of the the number number (81) (81)25.
(11)
Whi Which numb number er is lar larg ger (1. (1.2)4000 or 800 (9)
Answers :
(i)
18
(10)
1, 0 1 , 0 01
(1.2)4000.
(11)
Consider the expansion n
1
(x + y)n = (ii)
r%0
n
Cr x n r y r = nC x n y 0 + nC xn 1 y 1 + ...........+ nC xn r yr + ..........+ n C x 0 y n ....(i) 0 1 r n –
–
–
Now replace y 5 – y we get n
(x – y) n = (iii)
(iv)
1
r %0
n
Cr ( – 1) r x n r yr = nC x n y0 – nC x n 1 y1 + ...+ nC ( –1) r xn r yr + ...+ n C ( – 1) n x 0 yn ....(ii) 0 1 r n –
–
Adding (i) & (ii), we get (x + y) n + (x – y) n = 2[nC 0 xn y0 + nC2 x n
–
–
y2 +.........]
2
Subtracting (ii) from (i), we get (x + y)n – (x – y) n = 2[ nC1 xn 1 y1 + nC3 x n
–
–
3
y3 +.........]
(1 + x)n = C0 + C1x + C2x2 + ......... + Cr xr + .......... + Cnxn where Cr denotes nCr (1) (1)
......(1)
The sum sum of the the bin binomi omial al coef coeffi fici cien ents ts in the the exp expan ansi sion on of (1 + x)n is 2n Putting x = 1 in (1) n
C0 + nC1 + nC2 + ........+ nC n = 2 n
......(2)
n
or (2)
1
n
Cr % 2 n
r %0
Again putting x = –1 in (1), we get n
C 0 – nC1 + nC 2 – nC3 + ............. + ( –1) n nC n = 0
. . . .. . ( 3 )
n
or (3)
1 ('1)
r n
Cr % 0
r%0
The sum of of the binomia binomiall coeffi coefficie cients nts at odd odd posi positio tion n is equ equal al to to the the sum sum of the the binomi binomial al coeffi coefficie cients nts n 1 at even position and each is equal to 2 . from (2) and (3) –
n
(4)
C0 + nC2 + nC4 + ................ =
n
C1 + nC3 + nC5 + ................ = 2n
–
1
Sum Su m of of two two cons consec ecu utive tive bino binomi mial al coef coeffi fici cien ents ts n
Cr + nCr
–
=
1
=
n+1
Cr
0
L . H. S .
= nCr + nC r
–
1
=
n! n! + (n ' r )! r! (n ' r ! 1)! (r ' 1)!
n! n! (n ! 1)! 1 8 (n ! 1) ;1 ! = = = (n ' r )! (r ' 1)! 9: r n ' r ! 167 (n ' r )! (r ' 1)! r (n ' r ! 1) (n ' r ! 1)! r!
n+1
C r = R.H.S.
JEE(MAIN) BINOMIAL THEOREM - 7
Method : By Integration (1 + x)n = C0 + C1x + C2x2 + ...... + C n x n. Integrating both sides, within the limits – 1 to 0. 0
; (1 ! x )n ! 1 8 9 6 = 9: n ! 1 67 '1 1 n !1 C0 –
–
0
; x2 x3 x n!1 8 ! ..... ! Cn 9C0 x ! C1 ! C 2 6 2 3 n ! 167 9: '1
C1 C 2 C 8 ; ' ! ..... ! ( '1)n!1 n 6 0 = 0 – 9' C 0 ! 2 3 n ! 17 :
C2 C1 + 3 2
–
.......... + ( – 1) n
Cn 1 = Proved n !1 n !1
Example # 14 : If (1 + x) n = C 0 + C 1x + C2x2 + ........+ Cnxn, then prove that (i) C02 + C 12 + C 22 + ...... + C n2 = 2n C n ( i i) C0C 2 + C1C3 + C2C4 + .......... + C n 2 Cn = 2n Cn 2 or 2n Cn + 2 ( i i i) 1. C02 + 3 . C 12 + 5. C22 + ......... + (2n + 1) . Cn2 . = 2n. 2n 1C n + 2n Cn. Solution : (i) (1 + x)n = C0 + C1x + C2x2 + ......... + Cn x n. . .. . .. .. ( i) (x + 1)n = C 0xn + C 1xn 1+ C 2xn 2 + ....... + Cn x0 ........(ii) Multiplying Multiplying (i) an d (ii) (C 0 + C 1x + C2x2 + ......... + Cnxn) (C ( C0xn + C 1xn 1 + ......... + Cnx0) = (1 + x)2n Comparing coefficient of xn, C02 + C 12 + C 22 + ........ + Cn2 = 2n C n –
–
–
–
–
–
(ii) (ii)
From From the the pro produ duct ct of (i) (i) a and nd (ii) (ii) compa compari ring ng coef coeffi fici cien ents ts of xn C0C 2 + C1C3 + C2C4 + ........ + C n 2 C n = 2nCn 2 or 2nCn + 2. –
(iii)
–
or x n + 2 both sides,
2
–
Method : By Summation L.H.S. = 1. C02 + 3. C 12 + 5. C22 + .......... + (2n + 1) Cn2. n
1
=
r%0
(2r ! 1) nC 2 = r
n
n
1 2.r . ( C ) + 1 ( C ) n
n
2
r%0
2
r
r
r%0
n
=2
1. n .
n – 1
Cr
n –
1
Cr + 2n Cn
r %1
(1 + x)n = nC 0 + nC 1 x + nC2 x 2 + .............nCn x n ..........(i) (x + 1)n 1 = n 1C0 x n 1 + n 1C1 x n 2 + .........+n 1C n 1x 0 .........(ii) Multiplying Multiplying (i) and (ii) and comparing coeffcients of xn. n 1 C0 . nC 1 + n 1C 1 . nC 2 + ........... + n 1C n 1 . nCn = 2n 1Cn –
–
–
–
–
–
–
–
–
–
–
–
n
1
n'1
Cr '1 . nCr = 2n
–
1
Cn
r %0
Hence, required summation is 2n. 2n Method : By Differentiation
–
1
Cn + 2n Cn = R.H.S.
(1 + x2)n = C0 + C1x2 + C2x4 + C 3x6 + ..............+ Cn x 2n Multiplying both sides by x x(1 + x2)n = C0x + C1x3 + C2x5 + ............. + Cnx2n + 1 . Differentiating both sides x . n (1 + x2) n 1 . 2x + (1 + x2)n = C 0 + 3. C1x2 + 5. C2 x 4 + .....+ (2n + 1) Cn x 2n ......(i) (x 2 + 1) 1 )n = C0 x 2n + C1 x 2n 2 + C 2 x 2n 4 + ......... + C n ........(ii) Multiplying (i) & (ii) (C 0 + 3C 3 C1x2 + 5C 5 C2x4 + ......... + (2n + 1) Cn x 2n ) (C ( C0 x 2n + C 1x2n 2 + ........... + Cn) = 2n x2 (1 + x2)2n 1 + (1 + x 2)2n comparing coefficient of x2n, C02 + 3C 3 C12 + 5C 22 + .........+ (2n + 1) C n2 = 2n . 2n 1C n 1 + 2n Cn. –
–
–
–
–
–
–
–
C02 + 3C 3 C12 + 5C 22 + .........+ (2n + 1) C n2 = 2n . 2n 1Cn + 2nC n. Proved JEE(MAIN) BINOMIAL THEOREM - 9
As we know the Binomial Theorem – n
n
n
(x + y) =
1
n
Cr x
n –r
r
y
1 (n 'nr!)! r! x
=
n –r
yr
r%0
r %0
putting n – r = r 1 , r = r2 therefore,
1
n! (x + y) = r ! r2 ! r1 ! r2 % n 1 n
x r1 . y r2
Total number of term s in the expansion of ( x + y)n is equal to number of non-negative integral solution of r 1 + r 2 = n i.e. n+2 1C 2 1 = n+1 C1 = n + 1 –
–
In the same fashion we can write the multinomial theorem n! x 1r1 . x r22 ...x rkk (x 1 + x2 + x3 + ........... xk) = r ! r !... rk ! r1 ! r2 !...!rk %n 1 2
1
n
Here total number of terms in the expansion of (x1 + x2 + .......... + xk)n is equal to number of nonnegative integral solution of r 1 + r 2 + ........ + rk = n i.e. n+k 1C k 1 –
–
Example # 17 : Find the coefficient of a2 b3 c4 d in the expansion of (a – b – c + d) 10 Solution :
(a – b – c + d) 10 =
(10)! (a)r1 ( 'b)r2 ( 'c )r3 ( d)r4 r ! r ! r ! r ! 1 2 3 4 r1 !r2 !r3 ! r4 %10
1
we want to get a 2 b3 c 4 d this im plies that
<
(10 )! 3 4 – – – 2! 3! 4! 1! ( 1) ( 1) = 12600
coeff. of a2 b3 c4 d is
/ .
Example # 18 : In the expansion of -1 ! x ! 11
Solution :
7 , / -1 ! x ! * = x + .
r1 = 2, r2 = 3, r3 = 4, r4 = 1
1
11
7 , * , find the term independent of x. x +
(11)! r !r !r ! r1 ! r2 !r3 %11 1 2 3
r
/ 7 , 3 (1) ( x ) - * . x + r1
r2
The exponent 11 is to be divided among the b ase variables 1, x and
7 in such a way so that we x
get x0. Therefore, possible set of values of (r1, r2, r3) are (11, 0, 0), (9, 1, 1), (7, 2, 2), ( 5, 3, 3), (3, 4, 4), (1, 5, 5) Hence the required term is (11)! (11)! (11)! (11)! (11)! (11)! (7 0) + 7 1 + 7 2 + 7 3 + 7 4 + 7 5 9! 1 !1 ! 7! 2 ! 2 ! 5! 3 ! 3 ! 3! 4 ! 4 ! 1! 5 ! 5 ! (11)! =1+
(11)! 2! (11)! 4! (11) ! 6! . 7 1 + . 7 2 + . 7 3 9 ! 2 ! 1 ! 1! 7! 4! 2! 2! 5! 6! 3!3! +
(11) ! 8! (11) ! (10) ! . 7 4 + . 7 5 3!8! 4!4! 1 ! 10 ! 5!5!
= 1 + 11C2 . 2C1 . 71 + 11C4 . 4C2 . 72 + 11C 6 . 6C 3 . 73 + 11C8 . 8C 4 . 74 + 11C10 . 10 C5 . 75 5
=1+
1
r %1
11
C 2r . 2rCr . 7r
JEE(MAIN) BINOMIAL THEOREM - 11
Example-20 : If x is so small such that its square and higher powers ma y be neglected, then find the value of
(1 ' 3x )1 / 2 ! (1 ' x )5 / 3 ( 4 ! x )1 / 2 1 / 2
Solution :
(1 ' 3x )
! (1 ' x )
=
( 4 ! x )1 / 2
=
1'
5 / 3
3 5x x ! 1' 2 3
/ .
2-1 !
1 / 2
x , * 4 +
1 / 2 ' 19 x , / x , * -1 ! * = 6 + . 4 + 2 .
/ x 19 , x 1 / 2 ' 19 x , / 1 ' x , * * = 1 - 2 ' ' x * = 1 – 6 + . 8 + 8 2 . 2 . 4 6 +
–
'1 / 2
19 x = 1 – 12
41 24
x
Self practice problems : (16) (16) Find Find th the e possi possibl ble e set set of valu values es of of x for for whi which ch expa expansi nsion on of of (3 – 2x) 1/2 is valid in ascending powers of x. 2
(17)
3 1.3 / 2 , 2 1.3.5 / 2 , - * + ............., then find the value of y2 + 2y If y = + 2 ! - * + 5 5 3 ! . 5 + . +
(18)
T he co co ef f icient of of x100 in
(1 ' x )2 (B) –57
( A ) 10 0 Answers :
3 ' 5x
(16)
is
/ 3 3 , , * . 2 2 +
x & -'
(C) –197 (17)
4
(D) 53
(18)
C
JEE(MAIN) BINOMIAL THEOREM - 13
20.
x (1 x ' 1 # & Find the coefficient of the term independent of x in the expansion of $ 2 / 3 ' ! % x ' x1 / 3 ( 1 x ' x1 / 2 "
10
21.
If in the expansion of (1 + x)m (1 – x)n, the coefficients of x and x2 are 3 and –6 respectively. Then find the value of m.
22.
Find the number of terms in the expansion of (1 + 5 2 x)9 + (1 – 5 2 x)9.
23.
If the coefficients of second, third and fourth terms in the expansion of (1 + x)n are in A.P., A.P., then find the value of n.
24.
If in the expansion of (1 – x)2n –1 the coefficient of xr is denoted by ar, then prove that ar –1 + a2n –r = 0
25.
Using binomial theorem, prove that 23n – 7n – 1 is divisible by 49 where n ) N.
26.
Using binomial theorem, prove that 32n+2 – 8n – 9 is divisible by 64, n ) N.
27.
Prove that
28.
Find the sum of the infinite series 1 +
29.
Prove that (x2 – y2) +
1 1 1 1 & 4 # ( – – + ..... * = loge $ ! . 1.2 2.3 3.4 4.5 % e " 1 1 1 ( ( + ........ 2 ! 4! 6!
1 4 1 6 x2 y2 (x – y4) + (x – y6) + ...... to * = e – e 2! 3!
Type (IV) : Very Long Answer Type Questions: n
30.
Find the value of
+ ('1)
r n
Cr
r ,0
1 ( r loge 10 (1 ( loge 10 n )r
[06 Mark Each] .
31.
If the coefficient of rth, (r + 1)thand (r + 2)th terms in the expansion of (1 + x)14 are in A.P, then find the value of r.
32.
If the coefficients of three cosecutive terms term s in the expansion of (1 + x)n are in the ratio 1 : 7 : 42. Find n
33.
If 3rd, 4th, 5th and 6th terms in the expansion of (x + -)n be respectively a, b, c and d then prove that b 2 ' ac c 2 ' bd
=
5a 3c
34.
If coefficients of three consecutive terms in the expansion of (1 + x)n be 76,95 and 76. Then find n.
35.
If the 2nd, 3rd and 4th terms in the expansion of (x + a)n are 240, 720 and 1080 respectively, respectively, find x, a and n.
36.
Sum the series from n = 1 to n = * , whose nth term is (i)
37.
1 (n ( 1) !
1 (n ( 2) !
(iii)
1 (2n – 1) !
Prove that
& m # loge $ ! = 2 % n " 38.
(ii)
3& m – n # 1 & m – n #3 1 & m – n #5 0 1$ !( $ ! ( $ ! ( ..... 12% m ( n " 3 % m ( n " 5 % m ( n " ./
Prove that
3 1 0 1 1 & x ( 1 # ( ( ( ..... ! = 2 1 3 5 % x " 2 (2 x ( 1) 3(2x ( 1) 5(2x ( 1) /
loge $
JEE(MAIN) BINOMIAL THEOREM - 15
A-12.
The co-efficient of x in the expansion of (1 ' 2 x + 3 x 3
( 1) 56
( 2) 65
5
& 1 # ) $1 ( ! % x "
8
is :
( 3 ) 15 4
( 4) 62
5
A-13.
& 2 1 # The term containing x in the expansion of $ x ( ! is x " % (1) 2nd
A-14.
(2) 3rd
(3) 4th
(4) 5th
Given that the term of the expansion (x1/3 ' x'1/2 )15 which does not contain x is 5 m, where m) N,then m= (1) 1100 ( 2 ) 10 10 ( 3 ) 10 0 1 (4) none 4
A-15.
3
1 # & 1 # & The term independent of x in the expansion of $ x ' ! $ x ( ! is: x " % x " % (1) ' 3 ( 2) 0 (3) 1
(4) 3
10
A-16.
& x 3 # ( 2 !! isThe term independent of x in the expansion of $$ is% 3 2 x " (1) 3/2
(2) 5/4
(3) 5/2
(4) None of these 5
A-17.
& P ( Q # = ! % Q "
Let the co-efficients of xn in (1 + x) 2n & (1 + x)2n ' 1 be P & Q respe ctively, then $ (1) 9
A-18.
( 2) 27
( 3) 8 1
( 4 ) no n e o f t h e s e
If (1 + by)n = (1+ 8y + 24 y2 +....) where n)N then the value of b and n are respectively(1) 4, 2 (2) 2, – 4 (3) 2, 4 (4) – 2, 4 100
A-19.
52
The coefficient of x in the expansion (1) 100C47
A-20.
+
m ,0
100
Cm (x – 3) 100 –m. 2 m is :
(2) 100C48
(3) –100C52
(4) –100C100
The co-efficient of x 5 in the expansion of (1 + x)21 + (1 + x) 22 +....... + (1 + x) 30 is : (1) 51C5 (2) 9C5 (3) 31C6 ' 21C6 (4) 30C5 + 20C5 n
A-21.
& 1 # The term independent of x in (1 + x)m $1 ( ! is % x " (1) m – nCn
A-22.
(2) m + nCn
(3) m + 1Cn
(4) m + nCn+1
(1 + x) (1 + x + x2) (1 + x + x2 + x3)...... (1 + x + x 2 +...... + x100 ) when written in the ascending power of x then the highest exponent of x is (1) 5000 ( 2 ) 50 30 ( 3 ) 50 5 0 (4) 5040
Section (B) : Numerically greatest term, Remainder and Divisibility problems B-1.
The numerically greatest term in the expansion of (2 + 3 x) 9, when x = 3/2 is (1) 9C6. 29. (3/2)12 (2) 9C3. 29. (3/2)6 (3) 9C 5. 29. (3/2)10 (4) 9C4. 29. (3/2)8
B-2.
The numerically greatest term in the expansion of (2x + 5y) 34, when x = 3 & y = 2 is : (1) T21 (2) T22 (3) T23 (4) T24
B-3.
The remainder when 22003 is divided by 17 is : (1) 1 ( 2) 2 (3) 8
( 4 ) no n e o f t h e s e JEE(MAIN) BINOMIAL THEOREM - 17
C-9.
& 50 # !! % 0 "
The value of $$
& 50 # & 50 # $$ !! + $$ !! % 1 " % 1 "
& 100 # !! (1) $$ % 50 "
& 100 # !! (2) $$ 51 % " 10
C-10.
The value of
+
r ,1
(1) 5 (2n – 9)
C-11.
C-13.
n
Cr
Cr '1
& 50 # $$ !! is, where nCr = % 50 "
& n # $$ !! % r " 2
& 50 # (3) $$ !! % 25 "
& 50 # (4) $$ !! % 25 "
(3) 9 ( n – 4)
(4) none of these
is equal to
( 2) 1 0 n
+
10 & 10 # $ ( '1)K CK ! is : $ 2K "! % K ,0
+
(2) 220
& In the expansion of (1 + x)n $1 ( %
(4) 25
(3) 1 n
1 # ! , the term independent of x isx "
(1) C20 + 2 C12 +.....+ (n + 1) Cn2
(2) (C0 + C1 +....+ Cn)2
(3) C 20 + C12 +.......+ Cn2
(4) None of these
If (1 + x)n = C 0 + C 1x + C 2x2 +...+Cn.xn then for n odd, C12 + C 32 + C 52 +.....+ Cn2 is equal to (2n)! ( 2n)! (1) 22n – 2 (2) 2n (3) (4) 2 2(n! ) (n! ) 2 n
C-14.
n
r.
& 10 10 # $ Cr ! The value of the expression $ ! % r ,0 " (1) 210
C-12.
& 50 # & 50 # $$ !! +...........+ $$ !! % 2 " % 49 "
If a n = (1)
+ r ,0
n a 2 n
1 n
Cr
n
, the value of
(2)
+ n 'C2r is : r ,0
n
r
1 a 4 n
(3) nan
(4) 0
Section (D) : Multinomial Theorem, Binomial Theorem for negative and fractional index D-1.
The coefficient of a 5 b4 c7 in the expansion of (bc + ca ( ab) 8 is ( 1 ) 2 80 ( 2) 240 ( 3 ) 18 0
( 4) 32
D-2.
If 4 x4 < 1, then the co-efficient of xn in the expansion of (1 + x + x2 + x 3 +.......)2 is (1) n ( 2) n ' 1 (3) n + 2 (4) n + 1
D-3 D-3
The coefficient of x4 in the expression (1 + 2x + 3x2 + 4x3 + ......up to *)1/2 (where | x | < 1) 1) is (1) 1 (2) 3 (3) 2 (4) 5
Section (E) : Exponential and Logarithmic series E-1_.
Sum of the infinite series 1 1( 2 1( 2 ( 3 ( ( + ..... to * 2! 3! 4! (1)
E-2_.
e 3
( 2) e
(3 )
e 2
(4) none of these
(3)
2 45
(4) none of these
The coefficient of x6 in series e 2x is (1)
4 45
(2)
3 45
JEE(MAIN) BINOMIAL THEOREM - 19
4.
& 3 1 # In the expansion of $$ 4 ( 4 !! 6 " %
20
(1) (1) the num umbe berr of irra irrattiona ionall term terms s is 19 ( 3 ) t h e nu m b er of r a t io na l t e r m s is 2 5.
6.
(2) (2) mid middle dle term term is irrat rratiional onal ( 4 ) All of t he s e
If (1 + 2x + 3x 2)10 = a 0 + a 1x + a2x2 +.... + a20x20, then : (1) a1 = 20 ( 2) a2 = 210 (3) a4 = 8085
( 4 ) All of t h es e
(1 + x + x2 + x3)5 = a0 + a1x + a2x2 +....................... + a15 x15 , then a10 equals to : ( 1) 99 ( 2) 101 ( 3 ) 10 0 (4) 110 n
7.
& 3 1 # In the expansion of $ x ' 2 ! , n ) N, if the sum of the coefficients of x 5 and x10 is 0, then n is : % x " ( 1) 25
( 2) 20
( 3) 1 5
( 4) None o f thes e 10
8.
& # $ x (1 x '1 ! ' The coefficient of the term independent of x in the expansion of $ 2 is : 1 1 ! $ 3 ! % x ' x 3 ( 1 x ' x 2 " ( 1) 70
(2) 112
( 3 ) 10 5
( 4 ) 2 10
9.
T he ter m in in the exp ansion of (2 (2 x – 5)6 which has greatest binomial coefficient is (1) T3 (2) T4 (3) T5 (4) T6
10.
The remainder when 798 is divided by 5 is (1) 4 ( 2) 0
(3) 2
(4) 3
The last three digits of the number (27)27 is ( 1 ) 8 05 ( 2) 301
( 3 ) 50 3
( 4 ) 8 03
79 + 9 7 is divisible by : (1) 7 ( 2) 24
( 3) 6 4
( 4) 72
11.
12.
13.
Let f(n) = 10n + 3.4n +2 + 5, n ) N. The greatest value of the integer which divides f(n) for all n is : ( 1) 27 ( 2) 9 (3) 3 ( 4) None o f thes e
14.
Coefficient of x n ' 1 in the expansion of, (x + 3)n + (x + 3) n ' 1 (x + 2) + (x + 3) n ' 2 (x + 2)2 +..... + (x + 2)n is : (1) n+1C2(3) (2) n'1C2(5) (3) n+1C2(5) (4) nC2(5)
15.
The term in the expansion of (2x – 5)6 which has greatest numerical coefficient is (1) T3 ,T 4 (2) T4 (3) T5 , T6 (4) T 6 , T7
16.
Number of elements in set of value of r for which, 18Cr ' 2 + 2. 18Cr ' 1 + 18Cr 5 20C 13 is satisfied : (1) 4 elem ents ( 2 ) 5 e l e m en t s ( 3 ) 7 e l e m e nt s ( 4 ) 1 0 e lem en t s
17.
The number of values of ' r ' satisfying the equation, 39 C3r '1' 39C (1) 1
18.
( 2) 2
The sum
(1)
1 n!
1 1 ! ( n ' 1) !
(2n ' 1 ' 1) 1)
(
(3) 3
1 2 ! (n ' 2 ) !
(2)
2 n!
( ......
(2n ' 1) 1)
1 1 ! ( n ' 1) !
r2
= 39 Cr 2 '1' 39C 3r is : (4) 4
is equal to :
(3)
2 n!
(2n'1 ' 1)
(4) none
JEE(MAIN) BINOMIAL THEOREM - 21
PART - II : COMPREHENSION Comprehension # 1 Let P be a product given by P = (x + a1) (x + a2) ......... (x + an) n
and
Let S1 = a 1 + a 2 + ....... + an =
+ a , S = + + a .a , S = + ++ + + a .a .a i
i ,1
i
2
j
i 6 j
i
3
j
k
and so on,
i 6 j 6 k
then it can be shown that P = x n + S 1 xn – 1 + S2 x n – 2 + ......... + Sn. 1.
The coefficient of x8 in the expression (2 + x)2 (3 + x)3 (4 + x)4 must be (1) 26 (2) 27 (3) 28
(4) 29
2.
The coefficient of x19 in the expression (x – 1) (x – 22) (x – 32) .......... (x – 202) must be (1) 2870 (2) 2800 (3) –2870 (4) – 4100
3.
The coefficient of x98 in the expression of (x – 1) (x – 2) ......... (x – 100) must be (1) 12 + 22 + 32 + ....... + 1002 (2) (1 + 2 + 3 + ....... + 100)2 – (12 + 22 + 32 + ....... + 1002) 1 [(1 + 2 + 3 + ....... + 100) 2 – (12 + 22 + 32 + ....... + 1002)] 2 (4) None of these (3)
Comprehension # 2 We know that if nC0, nC1, nC2, ........., nCn be binomial coefficients, then (1 + x)n = C 0 + C 1 x + C 2 x2 + C 3x3 + ......+ C n x n . Various relations among binomial coefficients can be derived by putting
& $ %
x = 1, – 1, i, 7 $ where i ,
4.
i 3 #! . 2 "!
The value of nC0 – nC2 + nC4 – nC6 + ....... must be (1) 2i (2) (1 – i)n – (1 + i) n (3)
5.
1 2
' 1, 7 , ' (
1 [(1 – i)n + (1 + i)n] 2
(4)
1 [(2 – i)n + (1 – i)n] 2
The value of expression (nC0 – nC2 + nC4 – nC6 + .......)2 + (nC1 – nC3 + nC5 .........)2 must be (1) 22n
(2) 2n
(3) 2 n
2
(4) None of these
PART - I : AIEEE PROBLEMS (LAST 10 YEARS) 1.
If n
n
Cr denotes the number of combinations of n things taken r at a time, then the expression
Cr (1 ( nCr '1 ( 2 8 nCr equals
(1) n( 2 Cr 2.
(2) n( 2 Cr (1
[AIEEE 2003] (3) n(1Cr
The number of integral terms in the expansion of
9
(1) 32
(3) 34
(2) 33
3 ( 8 5
(4) n(1Cr (1
:
256
is :
[AIEEE 2003] (4) 35. JEE(MAIN) BINOMIAL THEOREM - 23
13.
The sum of the series 20C0 – 20C1 + 20C2 – (1) –20C10
14.
(2)
1 2
20
C3 + ..... + 20C10 is
20
C10
[AIEEE 2007 (3, –1), 120] (4) 20C10
(3) 0
a equals b [AIEEE 2008 (3, –1), 105]
In the binomial expansion of (a – b)n , n 5 5, the sum of 5th and 6th term is zero, then
(1)
n'4 5
(2)
5
(3)
n'4
6
(4)
n'5
n'5 6
n
15.
Statement-1 :
+ (r ( 1) C = (n + 2) 2 n
n –1
r
[AIEEE 2008 (3, –1), 105]
r ,0
n
Statement-2 :
+ (r + 1) r ,0
nC xr = r
(1 + x)n + nx (1 + x)n – 1
(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 Statement-2 is NOT a correct explanation for Statement-1 Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement-2 is True 10
16.
Let S1 =
+
j ( j – 1)
j , 1
10
10
C j , S2 =
+
j
j , 1
10
10
C j and S3 =
+ j
2 10
j , 1
C j.
[AIEEE 2009 (4, –1), 144]
Statement -1 : S3 = 55 × 2 9 . Statement -2 : S1 = 90 × 28 and S2 = 10 × 28. (1) Statement Statement -1 is true, Statement-2 Statement-2 is true ; Statement Statement -2 is not a correct explanation for Statement Statement -1. (2) Statement-1 Statement-1 is true, Statement-2 Statement-2 is false. (3) Statement Statement -1 is false, Statement Statement -2 is true. (4) Statement Statement -1 is true, Statement Statement -2 is true; Statement-2 Statement-2 is a correct explanation for Statement-1. Statement-1.
17.
The coefficient of x7 in the expansion of (1 – x – x2 + x3)6 is : (1) 144 (2) – 132 (3) – 144
18.
If n is a positive integer, then
9 3 ( 1:
(1) an irrational number (3) an even positive integer
2n
9 3 ' 1:
2n
–
[AIEEE 2011 (4, –1), 120] (4) 132
is :
[AIEEE-2012, (4, –1)/120]
(2) an odd positive integer (4) a rational number other than positive integers
10
19.
x (1 x ' 1 # & ' ! is : [AIEEE - 2013, (4, – ¼) 120 ] The term independent of x in expansion of $ 2 / 3 % x ' x1 / 3 ( 1 x ' x1 / 2 " (1) 4
(2) 120
(3) 210
(4) 310
JEE(MAIN) BINOMIAL THEOREM - 25
BOARD LEVEL SOLUTIONS
5.
Type (I) 1.
4
9 :
C0 x3
4
Puttin Putting g x = 2, we we get get
º
1
& 4 # & 4 # $ ! + 4 C1( x 3 )3 $& 4 #! + 4 C 2 ( x 3 )2 $ ! % x " % x " % x " 3
& 4 # 4 3 0 & 4 # + C 3 ( x ) $ ! + C 4 ( x ) $ ! % x " % x " 4
256
+ 0.088 + 0.025 < e = 7·378 < e2 = 7·4 (correct to one decimal place) 2
6. loge(1 + 3x + 2x2) = loge[(1 + 2x) (1 + x)] = loge(1 + 2x) + loge(1 + x) 1
º & b # & b # 6 6 6 C 0 (ax) $ ' ! + C1 (ax)5 $ – ! % x " % x "
3 0 (2x )2 1 1 3 4 2 x ( ( 2 x ) ( 2 x ) ( ... * – – 1 . = 2 3 4 12 ./
2
& b # & b # + 6C2 (ax)4 $ – ! + 6C3 (ax)3 $ – ! % x " % x " 4
& b # + 6C6 (ax)6 $ – ! % x "
2 2 2 2 3 2 4 2 5 2 6 27 ( ( ( ( ( e = 1 + ( +... 1! 2 ! 3 ! 4 ! 5 ! 6 ! 7 ! 2
Ans.
x4
& b # & b # + 6C4 (ax)2 $ – ! + 6C5 ax $ – ! % x " % x "
2
< e2 = 1 + 2 + 2 + 1·333 + 0.666 + 0.266
4
3 1
; x12 + 16 x8 + 96 x4 + 256 +
2.
x x2 x3 x 4 ( ( ( We have, e = 1 + + .... to * 1! 2 ! 3 ! 4 ! x
3
3 2
+ 1 x –
5
= 3x –
1 2 1 3 1 0 ( x ) ( ( x ) – ( x )4 ( ...* . 2 3 4 /
5 2 17 4 x + 3x3 – x + ... * 2 4
Type (II)
6
7. ;
5
C0 (1 + x)5 ( –x2)0 + 5 C1 (1 + x)4 ( –x2)1
= a6x6 – 6a5bx4 + 15a4b2x2 – 20a3b3 +
15a 2b 4 x
4
3.
4
2
6ab 5
–
x
0
4
& x # & a # & # ! $' ! + 4 C1 $ x ! $ a! $ x! $ a! % " % " % "
C0 $
2
2
3
+ 5 C 2 (1 + x)3 ( –x2)2 +
& # $' a ! $ x! % "
& x # & a # ! $' ! + C 4 $$ ! $ x! a % " % "
4
4
=
4.
!
x2 a2
4x a a2 + 6 – 4 + 2 – a x x
Ans.
3
+ 5[1 + 4x + 6x2 + 4x3 + x4]( –x2) + 10[1 + 3x + 3x2 + x3] (x4) + 10[1 + x2 + 2x] ( –x6) + 5(1 + x) (x8) + ( –x10) ; (1 + 5x + 10x2 + 10x3 + 5x4 + x5) + [ – –5x2 – 20x3 – 30 x4 – 20x5 – 5x6] + (10x4 + 30x5 + 30x6 + 10x7) + [ – –10 x6 – 10x8 – 20x7] + 5x8 + 5x9 – x10 ; –x10 + 5x9 – 5x8 – 10x7 + 15x6 + 11x5 – 15x4 – 10x3 + 5x2 + 5x + 1 Ans. 1
& 1 # & 1 # 8. ; C 0 (x + 2)3 $ ' ! + 3 C1 (x + 2)2 $ ' ! % x " % x " 3
2
3
3
Thus the coefficient of x2 in the expansion of e
; (1 + 5x + 10x2 + 10x3 + 5x4 + x5)
& 1 # & 1 # + C 2 (x + 2)1 $ ' ! + 3 C 3 (x + 2)0 $ ' ! % x " % x "
3
3
3 (2x ) (2x )2 (2x)3 0 ( ( ( .... = e .e = e 11 ( 1! 2! 3! 12 ./
2x+3
C3 (1 + x)2 ( –x2)3 + 5 C 4 (1 + x)1 ( –x2)4
º
e2x+3 2x
5
+ 5 C 5 (1 + x)0 ( –x2)5
& # $' a ! $ x! % " 1
0
x
+
6 1
& x # & a # & # ! $' ! + 4 C3 $ x ! + C 2 $$ ! $ ! $ a! % a " % x " % " 4
b6
22 is e = 2e3 2! 3
& 1 # ; [x3 + 8 + 12x + 6x2] + 3.[x2 + 4x + 4]. $ ' x ! % " & 1 # 1 ! 2 – % x " x 3
+ 3(x + 2). $
JEE(MAIN) BINOMIAL THEOREM - 27
n
16. As Tr+1 =
Cr x n'r y r in (x + y)n
7
& 1 # & 1 # 7 = C 0 $ ! + C1 $ ! % 2 " % 2 "
17
& 7 # Now consider $ x ' ! % x "
7
[on comparing n = 17, r = 10, x = x , y =
'7 x
10
T11 =
17
]
& 7 # ( '7 ) ! = 17 C10 x 7 . 10 % x " x
17
C10 x '3
Ans.
n
C 0 +
n
C 2 x 2 +.......+ n Cn x n ]
100
+
100
C0 +
100
C1 (100)1 100
C2 (100)2 +....... +
100
!
100
1+
C100 (100)100
C2 +....... 10196]
C2 +..... + 10196 is a natural number by the
;
C0 (7)
+
1995
1994
C1 (7)
+ ....
C1995 (10)
+.....
1995
C1995 (10)1995 – 71995
1995
1995
+
1995
C 0 +
1995
; 1 + 10N [! 1995
C1 +......... 1995
C1995 (10)1994 +
1995
C1 x
n –1
C1995 (10)1994]
y +
n
7
6
4x ( 1 + 2
1 26
7
& 1 # C3 $ ! % 2 "
4
& # $ 4x ( 1 ! $ 2 ! % "
5
3
7
º & # & # $ 4 x ( 1 ! + 7 C7 $& 1 #! $ 4 x ( 1 ! ] $ 2 ! % 2 " $% 2 "! % "
1 2
7
+
7
C3 .
1 27
. (4x + 1)
( 4 x ( 1)3 (4x + 1) + ] 27 27 1
2
7 4 x ( 1 [ C1 +
7
C3 (4x + 1)
+ 7 C5 (4x + 1)2 + (4x + 1)3] 7 7A D& # & # 1 4 x 1 1 4 x 1 ( ( ' ( > 1 ! '$ ! >@ C$$ ! $ ! 2 2 4x ( 1 > " % " >? B%
=
1 6
[ 7 C1 + 7 C3 (4x + 1) + 7 C5 (4x + 1)2 + (4x + 1)3]
2 < It is a polynomial polynomial of degree 3. Ans.
C1 (7)1994 +.......
1995
C1 +........
1995
C1995 (10)1994]
& # $ 1 ( 4 x ( 1 ! [! (x + y)n = n C xn + $2 0 2 "! % 1
C5 .
<
7
n
7
C1 (10)
= N(natural number as it is the sum of binomial coefficients) < Units place is 1 Ans.
19. Consider
2
1
C0 + 10
1995
& # $ ' 4x ( 1 ! + $ 2 "! %
2
7 = 2 4 x ( 1 [ C1 .
1
[ 1995 C1(7)1994 +........ 1995 C1995 (10)1994 +
& 1 # + C5 $ ! % 2 "
=
(10)
1995
; Now
6
& # & # $ ' 4 x ( 1 ! +...+ 7 C7 $ ' 4 x ( 1 ! ...(ii) $ $ 2 "! 2 "! % %
7
+
18. Consider 171995 + 111995 – 71995 ; (7 + 10)1995 + (1 + 10)1995 – 71995 1995
5
& 1 # = 2[ C1 $ ! % 2 "
virtue of its being the binomial coefficients. = 104 N < (101)100 – 1 is divisible by 10,000.
1995
& 1 # C2 $ ! % 2 "
7
Now (1+ 100)100 – 1 = 1 + 104 + 100 C2 104 +...10200 – 1 = 104 [1 +
7
(i) – (ii)
Using Binomial theorem (1 + 100)100 =
7
7
7
C1 x +
2
& 1 # & 1 # 7 = C 0 $ ! + C1$ ! % 2 " % 2 "
17. (101)100 = (1 + 100)100 n
4x ( 1 + 2
& # & # $ 4 x ( 1 ! +.....+ 7 C7 $ 4 x ( 1 ! ...(i) $ 2 ! $ 2 ! % " % "
5
7
Type (III)
[(1 + x)n =
& 1 # C2 $ ! % 2 "
& 1 4 x ( 1 #! Now $ ' $2 2 "! %
10
C10 (x)17 –10 $ '
< T11 = 710
6
7
n –2
2
n
n
C 2 x y +....... + Cn y ]
20. Let = x = t6 6 & 6 # $ t (1 ' t '1 ! $ t4 ' t2 ( 1 t6 ' t3 ! % "
10
3 ( t 2 ( 1) ( t 4 ' t 2 ( 1) ( t 3 ' 1) ( t 3 ( 1) 0 ' . ; 1 t 4 ' t2 ( 1 t 3 ( t 3 ' 1) ./ 12 3 t 5 ( t 3 ' t 3 ' 10 . ; 1 t3 21 /.
10
10
JEE(MAIN) BINOMIAL THEOREM - 29
=
(2n ' 1)! (2n ' 1)! ( 1)r –1 + ( 1)2n –r – (2n ' 1)! (n ' 1)! (r ' 1)! ( 2n ' r )! –
( 2n ' 1)! = (2n ' r )! (r ' 1)! [( –1)r –1 + ( –1)2n –r ] (2n ' 1)! = (2n ' r )! (r ' 1)!
& 4 # % "
3 ' 1r 1 0 ( 1 . 12 ' 1 ( '1)r ./
C 0 +
n
Put x = 1, we get
C1x + ....... +
n
1
n
2
n
Cn x n ]
n
= n C 2 (7)2 ( nC3 (7)3 ( ...... ( nCn (7)n
= 49[ n C 2 ( nC 3 7 ( ...... ( nCn 7 n'2 ]
3
e + e –1 = 2 11 (
2
& 2 y 4 y 6 # ( ( ...... !! – $$ y ( 2! 3 ! % "
It is a natural number by the virtue of being a sum of binomial coefficients. 23n – 7n – 1 = 49 N < 23n – 7n – 1 is divisible by 49. Proved.
& # ( x E ) 2 ( x 2 )3 2 $ ! 1 ( x ( ( ( ... = $ ! 2 ! 3 ! % "
26. Consider 32n+2 – 8n – 9 = (32)n+1 – 8n – 9 = (9)n+1 – 8n – 9 = (1 + 8)n+1 – 8n – 9 ... same
& # ( y 2 )2 ( y 2 )3 2 $ ! 1 ( y ( ( ( ... – $ ! 2 ! 3 ! % "
C0 ( n(1C1(8)1 ( n (1C2 (8)2 ( ....... ( n(1Cn (1(8)n(1 ' 8n ' 9
= 1 + (n + 1) 8 +
n (1
C 2 (8) ( ....... ( 8 2
n(1
& = $ 1 – %
2
= e x – e y
' 8n ' 9
= 1 + 8n + 8 + n+1C2 (8)2 + .......+ 8n+1 – 8n – 9 = n+1C2 (8)2 + n+1C3 (8)3 + ....... + 8n+1 = (8)2 [n+1C2 + n+1C3 (8) + ........ + 8n –1 ] n+1 C2 + n+1C3 (8) + ...... + 8n –1 = N It is a natural number by the virtue of being a sum of binomial coefficients. < 32n+2 – 8n – 9 = 64N < 32n+2 – 8n – 9 is divisible by 64 27. L.H.S. =
0 1 1 1 ( ( ( ..... * . 2! 4 ! 6 ! /
1 1 1 ( ( + .... * = 1 (e + e –1) 2! 4! 6! 2
Hence 1 (
C 2 ( nC 3 .7 ( ...... ( nCn 7n'2 = N
n(1
1 1 1 ( – +....... * 1! 2! 3 !
& 2 x 4 x 6 x 8 # $ ( ( ( ......!! 29. L.H.S. = $ x ( 2! 3 ! 4 ! % "
= 7 2 [ nC 2 ( nC 3 7 ( ...... ( nCn 7 n'2 ]
=
1 1 1 ( ( + .... * 1! 2! 3 !
add both equation, we get
n
= 1 ( 7n ( nC 2 (7)2 ( nC3 (7)3 ( ... ( nCn (7)n ' 7n ' 1
n
e=1+
Put Put x = – 1, we get e –1 = 1 –
= C 0 ( C1(7) ( C 2 (7) ( ...... ( Cn (7) ' 7n ' 1 n
x x2 x3 ( ( +..... to * 1! 2 ! 3 !
28. We have ex = 1 +
25. Consider 23n – 7n – 1 = (8)n – 7n – 1 = (1 + 7)n – 7n – 1 n
1 1 1 1 0 ( – ( ......to * . – 1 2 3 4 5 /
= 2 loge 2 – 1 = loge4 – logee = loge $ e ! = R.H.S.
(2n ' 1)! = ( 2n ' r )! (r ' 1)! [0] = 0 proved.
[! (1 + x)n =
3 2
= 2 11 –
Type (IV) 30. Let loge 10 = x n
Now
+ ('1)
r n
;
Cr
r ,0
n
+
( '1)r nCr
r ,0
1 1 1 1 ( – – + ..... to * 1.2 2.3 3.4 4.5
1 # & 1 1 # & 1 1 # & 1 1 # ! – $ – ! ( $ – ! – $ – ! + ...... 2 " % 2 3 " % 3 4 " % 4 5 "
2
1( r x
1 (1 ( xn)r
n
;
+ ('1)
r n
r ,0
Cr
[! log am = m log a]
(1 ( xn)r
n
+
+ ('1)
r n
r ,0
& 1 1 1 1 # = 1 – 2 $ – ( – .....to * ! % 2 3 4 5 "
rx (1 ( nx)r
r (1 ( nx )r n
1 1 1 1 1 1 1 1 = 1 – – ( ( – – ( ( .......... 2 2 3 3 4 4 5 5
Cr
+
n x ( '1)r + r 1 ( nx r ,1
n '1
[using n Cr =
r
Cr '1
n r
(1 ( nx )r '1
n '1
Cr '1 ]
JEE(MAIN) BINOMIAL THEOREM - 31
n 2 n n b 2 ' ac x 2n ' 6 - 6 [( C 3 ) ' C 2 . C 4 ] v Now = 2 = 2n' 8 8 n 2 n n vi - [( C 4 ) ' C3 . C5 ] c ' bd x
35. As Tr+1 = n Cr xn –r yr in (x + y) n & consider (x + a)n
3 0 n! n! n! n! ' 1 . x 2 2 (n ' 3 )! (n ' 3 )!3 ! 3 ! (n ' 2)! 2 ! (n ' 4)! 4 ! / ; 2 0 n! n! n! n! - 3 1 (n ' 4)! (n ' 4)! 4 ! 4 ! ' (n ' 3)! 3 ! . (n ' 5)! 5 ! . 2 /
3 1 1 0 ' 1 . x 2 (n ' 3)3 4(n ' 2) / 3 1 n ! n! 1 0 1 -2 ' 8 1 . (n ' 4)! (n ' 5)! 2 (n ' 4)4 5(n ' 3) / 3 ! 4 !
;
=
-
2
n
Cr x r in [1 + x]n
4.
Cr (1 x r (1
Cr '1 = 76
...(i)
n
Cr = 95
...(ii)
Cr (1 = 76
...(iii)
Now
ii = i
n
Cr
Cr '1
=
95 n'r (1 = 76 r
; 76n – 76r + 76 = 95 r ; 76(n + 1) = 101 r ...(iv) iii = ii
n
n
C3
a 9 = x 2
...(v)
6 4 n '1 = .2 = 9 3 n'2
a a 3 3x = 6 ; = ; a = x x 2 2 3x in (i) ; n C1 xn –1 a1 = 240 2
3x = 240 2 x5 = 32 ; x = 2
5.x4.
3x = 3 2 < n = 5; a = 3; x = 2
Now a =
n
Cr
*
(i)
Sum =
76 n'r = = 95 r (1
95n – 95r = 76r + 76 95n – 76 = 101r ...(v) From (iv) 95n – 76 = 76n + 76 19n = 152 n=8 Ans.
Ans.
36.
+t
n
n ,1
= t1 + t2 + t3 + ..... to *
*
Cr (1
...(iv)
3 1080 x n'3 a 3 . n'2 . 2 = = n C2 x 720 2 a
Put a =
n
n
C 2 x n '2 a 2 . = 3 n C1 x n'1 a1
3n – 3 = 4n – 8 n=5 From (iv)
Now it is given that coefficients of Tr , Tr+1 and Tr+2 are 76, 95, 76 repsectively.
n
n
n (n ' 1) a a . = 3 ; (n – 1). = 6 2n x x
9iv : 9v :
< Tr = n Cr '1 x r '1 ; Tr+1 = n Cr x r Tr+2 =
...(iii)
(n – 2)
34. Let Tr , Tr+1 and Tr+2 be the three consecutive terms in the expansion of (1 + x)n n
T4 = n C 3 xn –3 a3 = 1080
n (n ' 1) (n ' 2).2 a 3 = 6n (n ' 1) x 2
(n ' 4)! 20 (n ' 2)!
As Tr+1 =
...(ii)
9iii: 9ii: ;
x 2 4 ! (n ' 5)! 3 (n ( 1) 4.5(n ' 4) 0 8 ; 2. 1 . (n ( 1) / - 2! (n ' 3)! 212.(n ' 2) x2
T3 = n C 2 xn –2 a2 = 720
=
x 2 4 ! (n ' 5)! 3 4n ' 8 ' 3n ( 9 5(n ' 3 ) (n ' 4 ) 0 . 8 1 . 2 - 2 ! (n ' 3)! 2 3.4 (n ' 2).(n ' 3) 5n ' 15 ' 4n ( 16 /
;
(given) ...(i)
9ii: 9i: ;
1 n! n! 2! 3 ! (n ' 3)! (n ' 4)!
2
T2 = n C1 xn –1 a1 = 240
=
+ (n (11) ! = n ,1
1 1 1 ( ( + ..... to * 2! 3! 4 !
3& 1 1 1 # 0 $ !! – 2. = e – 2 ( ( ( * 1 ..... 1 = $ " / 2% 1! 2 ! 3 ! (ii) We hav have, tn =
1 (n ( 2) !
*
Sum =
+ (n (12) ! n ,1
JEE(MAIN) BINOMIAL THEOREM - 33
OBJECTIVE QUESTIONS * Marked Questions may have more than one correct option. 1.
If the sum of the co-efficients in the expansion of (1 + 2x)n is 6561, then the greatest term in the expansion for x = 1/2 is : (1) 4th (2) 5th (3) 6th (4) none of these 6
2.
6 & # 2 2 2 & # $ ! The expression, $% 2 x ( 1 ( 2x ' 1 "! ( $ ! is a polynomial of degree 2 2 ( ( ' 2 x 1 2 x 1 % "
(1) 5 3.
4.
(2) 6
+
n
n
C15 '3r Cr
r ,0
+
C 5r
(3)
r ,0
(2) r 5 (n ' 2) / 2
+
n
3
C 3r
(4)
r ,0
+
n
n
C 3 ' r C 5r
r ,0
91 ( x :n 1' x
is 2n, (|x| < 1), then –
(3) r F (n ( 2) / 2
(4) r 5 n
The coefficient of xn in polynomial (x + 2n+1C0) (x + 2n+1C1)........(x + 2n+1Cn) is (1) 2n + 1 (2) 22n+1 – 1 (3) 22n (4) none of these
& r '1 # $ n C r C 2p ! is equal to r p $ ! p 0 , % "
+ + r ,1
(1) 4n – 3n + 1 n
(2) 4n – 3n – 1
(3) 4n – 3n + 2
(4) 4n – 3n
C0 – 2.3 nC1 + 3.32 nC2 – 4.33 nC3 +..........+ ( –1)n (n +1) nCn 3n is equal to n
& 3n # ( 1! % 2 "
n (1) 9' 1: 2 $
9.
(2)
5
n
If n is even natural and coefficient of x in the expansion of
n
8.
5
r
(1) r F n / 2
7.
(4) 60
Co-efficient of x15 in (1 + x +x3 + x4)n is : (1)
6.
(4) 8
Co-efficient of x5 in the expansion of (1 + x2)5 (1 + x)4 is : (1) 40 (2) 50 (3) 30
5
5.
(3) 7
& %
n (2) 2 $ n (
3 # ! 2 "
(3) 2n + 5n 2n
(4) ( –2)n.
If the sum of the coefficients coef ficients in the expansion of (2 + 3cx + c2x2)12 vanishes, then c equals to (1) –1, 2 (2) 1, 2 (3) 1, –2 (4) –1, –2 4
10.
1 # & 2 ! is The term independent of x in the expansion of ( 1 + x + 2x ) $ 3 x ' 3 x 2 " % 2
(1) 10 11*.
12.
(2) 2
(3) 0
1000n Let an , for n ) N, then an is greatest, when n! (1) n = 997 (2) n = 998 (3) n = 999
& n # & n # & n # 2k $$ !! $$ !! – 2k '1 $$ !! % 0 " % k " % 1 " (1) nCk
& n ' 1 # $$ !! + 2k '2 % k ' 1 " (2) n+1Ck
& n # $$ !! % 2 "
& n # & n ' 2 # $$ !! –...... + ( – 1) k $$ k !! % " % k ' 2 " (3) n –1Ck
(4) 6
(4) n = 1000
& n ' k # $$ !! = % 0 " (4) n+2Ck JEE(MAIN) BINOMIAL THEOREM - 35
