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Binomial Theorem “Obvious” is the most dangerous word in mathematics......... Bell, Eric Temple
Binomial expression : Any algebraic expression which contains two dissimilar terms is called binomial expression. 1
For example : x + y, x 2y +
xy
2
, 3 – x,
x2 1 +
1 3
( x 1)1/ 3
etc.
Terminology used in binomial theorem : 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
Note : n! = n . (n – 1)! ;
n
N
Mathematical meaning of nC r : The term nCr denotes number of combinations of r things choosen from n distinct things mathematically, nCr =
n! , n N, r W, 0 r n (n r )! r!
n Note : Other symbols of of nCr are and C(n, r). r n Properties related to C r :
(i)
n
Cr = nCn – r
Note : If nCx = nCy (ii)
n
Cr + Cr – 1 = n
(iii)
Either x = y or
n
n
Cr
=
Cr 1
n r
n+1
x+y=n
Cr
nr 1 r
(v)
If n and r are relatively prime, 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 = ............. =
Statement of binomial theorem : (a + b)n = nC0 anb0 + nC1 an–1 b1 + nC2 an–2 b2 +...+ nCr an–r br +...... + nCn a0 bn where n N n
or
n
(a + b) =
n
C r a n r b r
r0
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
n
(1 + x) =
n
Cr x r
r 0
JEE(MAIN) BINOMIAL THEOREM - 1
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Regarding Pascal’s Triangle, we note the following : (a) Each row of the triangle begins with 1 and ends with 1. (b) Any entry in a row is the sum of two entries in the preceding row, one on the immediate left and the other on the immediate right. Example # 3 : The number of dissimilar terms in the expansion of (1 – 3x + 3x 2 – x 3)20 is (A) 21 (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 – x 3)20 is 61.
General term : (x + y)n = nC0 x n y0 + nC1 x n–1 y1 + ...........+ nCr x n–r yr + ..........+ nCn x 0 yn (r + 1)th term is called general term and denoted by T r+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. 28th term of (5x + 8y)30
Example # 4 : Find
(i)
Solution :
(i)
T 27 + 1 = 30C27 (5x)30– 27 (8y)27 =
(ii)
4x 5 7th term of 5 2x T6 + 1
4x = C6 5
96
9
n
Cn – r + 1 x r – 1 yn – r + 1
4x 5 7th term of 5 2x
(ii)
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 :
The general term in the expansion of 91/ 4 81/ 6
1000 r
Tr+1
1 4 1000 = Cr 9
1000
is
r
1 8 6 =
1000
Cr 3
1000 r 2
r
22
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
Middle term(s) : (a)
n 2 If n is even, there is only one middle term, which is 2
(b)
n 1 If n is odd, there are two middle terms, which are 2
th
th
term.
n 1 1 and 2
th
terms.
Example # 6 : Find the middle term(s) in the expansion of 14
(i)
2 1 x 2
(ii)
3 3a a 6
9
JEE(MAIN) BINOMIAL THEOREM - 3
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Case -
(i) (ii) (iii)
When
n 1 a 1 b
is an integer (say m), then
T r+1 > T r when r < m (r = 1, 2, 3 ...., m – 1) i.e. T 2 > T 1, T 3 > T 2, ......., T m > T m–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
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 a 1 b
is not an integer (Let its integral part be m), then
T r+1 > T r
i.e. (ii)
r<
n 1 a 1 b
(r = 1, 2, 3,........, m–1, m)
T 2 > T 1 , T 3 > T 2, .............., T m+1 > T m
T r+1 < T r
i.e.
when
when r >
n 1 a 1 b
(r = m + 1, m + 2, ..............n)
T m+2 < T m+1 , T m+3 < T m+2 , .............., T n +1 < T n
Conclusion : When
n 1 a 1 b
is not an integer and its integral part is m, then T m+1 will be the numerically greatest
term. Note : (i)
In any binomial expansion, the middle term(s) has greatest binomial coefficient. In the expansion of (a + b)n If n No. of greatest binomial coefficient Greatest binomial coefficient n Even 1 Cn/2 n Odd 2 C(n – 1)/2 and nC(n + 1)/2 (Values of both these coefficients are equal ) (ii) In order to obtain the term having numerically greatest coefficient, put a = b = 1, and proceed 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 Tr + 1 Tr 15 Cr 315 – r (| – 5x|)r 15Cr – 1 315 – (r – 1) (|– 5x|)r – 1
3. 15 )! 15 )! |– 5x | (15 r ) ! r ! (16 r ) ! (r 1) ! 1 (16 – r) 3r 5 16 – r 3r 4r 16 r4
5.
JEE(MAIN) BINOMIAL THEOREM - 5
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Self practice problems : (8)
If n is a positive integer, then show that 32n + 1 + 2n + 2 is divisible by 7.
(9)
What is the remainder when 7103 is divided by 25 .
(10)
Find the last digit, last two digits and last three digits of the number (81)25.
(11)
Which number is larger (1.2)4000 or 800 Answers :
(9)
18
(10)
1, 01, 001
(11)
(1.2)4000.
Some standard expansions : (i)
Consider the expansion n
n
(x + y) =
r 0
(ii)
n
Cr x n–r yr = nC x n y0 + nC x n–1 y1 + ...........+ nC x n–r yr + ..........+ nC x 0 yn ....(i) 0 1 r n
Now replace y – y we get n n
(x – y) =
r 0
n
Cr (– 1) r x n–r yr = nC x n y0 – nC x n–1 y1 + ...+ nC (–1)r x n–r yr + ...+ nC (– 1)n x 0 yn ....(ii) 0 1 r n
(iii)
Adding (i) & (ii), we get (x + y)n + (x – y)n = 2[nC0 x n y0 + nC2 x n – 2 y2 +.........]
(iv)
Subtracting (ii) from (i), we get (x + y)n – (x – y)n = 2[nC1 x n – 1 y1 + nC3 x n – 3 y3 +.........]
Properties of binomial coefficients : (1 + x)n = C0 + C1x + C2x 2 + ......... + Cr x r + .......... + Cnx n where Cr denotes nCr (1)
......(1)
The sum of the binomial coefficients in the expansion of (1 + x)n is 2n Putting x = 1 in (1) n
C0 + nC1 + nC2 + ........+ nCn = 2n
......(2)
n
or
n
Cr 2n
r 0
(2)
Again putting x = –1 in (1), we get n
C0 – nC1 + nC2 – nC3 + ............. + (–1)n nCn = 0
......(3)
n
or
(1)
r n
Cr 0
r0
(3)
The sum of the binomial coefficients at odd position is equal to the sum of the binomial coefficients at even position and each is equal to 2n–1. from (2) and (3) n
(4)
C0 + nC2 + nC4 + ................ = nC1 + nC3 + nC5 + ................ = 2n–1
Sum of two consecutive binomial coefficients n
Cr + nCr–1 =
n+1
Cr
L.H.S.
= nCr + nCr–1 =
n! n! + (n r )! r! (n r 1)! (r 1)!
n! n! (n 1)! 1 (n 1) 1 = (n r )! (r 1)! = (n r )! (r 1)! r(n r 1) = (n r 1)! r! = r n r 1
n+1
Cr = R.H.S.
JEE(MAIN) BINOMIAL THEOREM - 7
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Method : By Integration (1 + x)n = C0 + C1x + C2x 2 + ...... + Cn x n. Integrating both sides, within the limits – 1 to 0. 0
0
(1 x )n 1 x2 x3 x n1 C2 ..... Cn = C 0 x C1 2 3 n 1 n 1 1 1 C1 C 2 C 1 ..... ( 1)n1 n – 0 = 0 – C 0 2 3 n 1 n 1
C0 –
C2 C1 Cn 1 + – .......... + (– 1)n = Proved 3 2 n 1 n 1
Example # 14 : If (1 + x)n = C0 + C1x + C2x 2 + ........+ Cnx n, then prove that (i) C02 + C12 + C22 + ...... + Cn2 = 2nCn (ii) C0C2 + C1C3 + C2C4 + .......... + Cn – 2 Cn = 2nCn – 2 or 2nCn + 2 (iii) 1. C02 + 3 . C12 + 5. C22 + ......... + (2n + 1) . Cn2 . = 2n. 2n – 1Cn + 2nCn. Solution : (i) (1 + x)n = C0 + C1x + C2x 2 + ......... + Cn x n. ........(i) (x + 1)n = C0x n + C1x n – 1+ C2x n – 2 + ....... + Cn x 0 ........(ii) Multiplying (i) and (ii) (C0 + C1x + C2x 2 + ......... + Cnx n) (C0x n + C1x n – 1 + ......... + Cnx 0) = (1 + x)2n Comparing coefficient of xn, C02 + C12 + C22 + ........ + Cn2 = 2nCn (ii)
From the product of (i) and (ii) comparing coefficients of x n – 2 or x n + 2 both sides, C0C2 + C1C3 + C2C4 + ........ + Cn – 2 Cn = 2nCn – 2 or 2nCn + 2.
(iii)
Method : By Summation L.H.S. = 1. C02 + 3. C12 + 5. C22 + .......... + (2n + 1) Cn2. n
=
n
n
(2r 1) nC 2 = r
r0
2.r . (nCr)2 +
r 0
(
n
Cr )2
r0
n
=2
. n .
n–1
Cr – 1 nCr + 2nCn
r 1
(1 + x)n = nC0 + nC1 x + nC2 x 2 + .............nCn x n ..........(i) (x + 1)n – 1 = n – 1C0 x n – 1 + n – 1C1 x n – 2 + .........+n – 1Cn – 1x 0 .........(ii) Multiplying (i) and (ii) and comparing coeffcients of x n. n–1 C0 . nC1 + n – 1C1 . nC2 + ........... + n – 1Cn – 1 . nCn = 2n – 1Cn n
n1
Cr 1 . nCr = 2n – 1Cn
r 0
Hence, required summation is 2n. 2n – 1Cn + 2nCn = R.H.S. Method : By Differentiation (1 + x 2)n = C0 + C1x 2 + C2x 4 + C3x 6 + ..............+ Cn x 2n Multiplying both sides by x x(1 + x 2)n = C0x + C1x 3 + C2x 5 + ............. + Cnx 2n + 1. Differentiating both sides x . n (1 + x 2)n – 1 . 2x + (1 + x 2)n = C0 + 3. C1x 2 + 5. C2 x 4 + .....+ (2n + 1) Cn x 2n......(i) (x 2 + 1)n = C0 x 2n + C1 x 2n – 2 + C2 x 2n – 4 + ......... + Cn ........(ii) Multiplying (i) & (ii) (C0 + 3C1x 2 + 5C2x 4 + ......... + (2n + 1) Cn x 2n) (C0 x 2n + C1x 2n – 2 + ........... + Cn) = 2n x 2 (1 + x 2)2n – 1 + (1 + x 2)2n comparing coefficient of x2n, C02 + 3C12 + 5C22 + .........+ (2n + 1) Cn2 = 2n . 2n – 1Cn – 1 + 2nCn. C02 + 3C12 + 5C22 + .........+ (2n + 1) Cn2 = 2n . 2n–1Cn + 2nCn. Proved JEE(MAIN) BINOMIAL THEOREM - 9
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Multinomial theorem : As we know the Binomial Theorem – n
n
(x + y)n =
n
Cr x n–r yr
=
n!
(n r )! r!
x n–r yr
r0
r 0
putting n – r = r1 , r = r2 therefore,
n! r ! r2 ! n 1
x r1 . y r2
(x + y)n =
r1 r2
Total number of terms in the expansion of (x + y)n is equal to number of non-negative integral solution of r1 + r2 = n i.e. n+2–1C2–1 = n+1C1 = n + 1 In the same fashion we can write the multinomial theorem (x 1 + x 2 + x 3 + ........... x k)n =
r1 r2 ...rk
n! x r1 . x r22 ...x rkk r ! r !... rk ! 1 n 1 2
Here total number of terms in the expansion of (x 1 + x 2 + .......... + x k)n is equal to number of nonnegative integral solution of r1 + r2 + ........ + rk = n i.e. n+k–1Ck–1 Example # 17 : Find the coefficient of a2 b3 c 4 d in the expansion of (a – b – c + d)10 Solution :
(a – b – c + d)10 =
r1 r2 r3 r4 2
(10 )! r1 r2 r3 r4 r ! r ! r ! r ! (a) ( b) ( c ) (d) 10 1 2 3 4
we want to get a b c 4 d this implies that
3
coeff. of a2 b3 c 4 d is
r1 = 2, r2 = 3, r3 = 4, r4 = 1
(10 )! 3 4 2! 3! 4! 1! (–1) (–1) = – 12600 11
7 Example # 18 : In the expansion of 1 x , find the term independent of x. x Solution :
7 1 x x
11
=
r1 r2 r3
(11)! r ! r !r ! 11 1 2 3
r
73 (1)r1 ( x )r2 x
The exponent 11 is to be divided among the base variables 1, x and
7 in such a way so that we x
get x 0. 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)! (70) + 9! 1 !1 ! 71 + 7! 2 ! 2 ! 72 + 5! 3 ! 3 ! 73 + 3! 4 ! 4 ! 74 + 1 ! 5 ! 5 ! 75 (11)! (11)! 2! (11)! 4! (11) ! 6! = 1 + 9 ! 2 ! . 1 ! 1 ! 71 + 7 ! 4 ! . 2 ! 2 ! 72 + 5 ! 6 ! . 3 ! 3 ! 73 (11) ! 8! (11) ! (10) ! + 3 ! 8 ! . 4 ! 4 ! 74 + 1 ! 10 ! . 5 ! 5 ! 75 = 1 + 11C2 . 2C1 . 71 + 11C4 . 4C2 . 72 + 11C6 . 6C3 . 73 + 11C8 . 8C4 . 74 + 11C10 . 10C5 . 75 5
=1+
11
C 2r . 2rCr . 7r
r 1
JEE(MAIN) BINOMIAL THEOREM - 11
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Example-20 : If x is so small such that its square and higher powers may be neglected, then find the value of
(1 3x )1/ 2 (1 x )5 / 3 ( 4 x )1/ 2
Solution :
(1 3x )1/ 2 (1 x )5 / 3 ( 4 x )1/ 2
=
3 5x x 1 1 / 2 x 1 2 19 x 2 3 1 = 1/ 2 6 4 2 x 21 4
1 =
x 1 2 19 x 1 x 1 2 x 19 x 19 41 = =1– – x =1– x 6 8 4 6 8 2 2 12 24
Self practice problems : (16) Find the possible set of values of x for which expansion of (3 – 2x)1/2 is valid in ascending powers of x. 2
(17)
3 1.3 2 2 1.3.5 2 If y = + 2! + + ............., then find the value of y2 + 2y 5 3! 5 5
(18)
The coefficient of x 100 in
(1 x )2 (B) –57
(A) 100 Answers :
3 5x
(16)
is
3 3 x , 2 2
(C) –197 (17)
4
(D) 53
(18)
C
JEE(MAIN) BINOMIAL THEOREM - 13
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20.
x 1 x 1 Find the coefficient of the term independent of x in the expansion of 2 / 3 1/ 3 x 1 x x 1/ 2 x
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., 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
r n
(1)
Cr
r 0
1 r loge 10 (1 loge 10n )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 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, 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)
m – n 1 m – n 3 1 m – n 5 .... mn 3mn 5mn
Prove that
1 1 1 x 1 .... = 2 loge 3 5 5(2x 1) x (2 x 1) 3(2x 1) JEE(MAIN) BINOMIAL THEOREM - 15
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A-12.
The co-efficient of x in the expansion of (1 2 x 3 + 3 x 5) 1 (1) 56
(2) 65
1 x
8
is :
(3) 154
(4) 62
5
A-13.
A-14.
2 1 The term containing x in the expansion of x is x nd rd (1) 2 (2) 3 (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) 1010 (3) 1001 (4) none 4
A-15.
(1) 3 A-16.
3
1 1 The term independent of x in the expansion of x x is: x x (2) 0
(3) 1
x 3 2 The term independent of x in the expansion of 3 2x
(1) 3/2
(2) 5/4
(4) 3 10
is-
(3) 5/2
(4) None of these 5
A-17.
P Q Let the co-efficients of x n in (1 + x)2n & (1 + x)2n 1 be P & Q respectively, then = Q (1) 9 (2) 27 (3) 81 (4) none of these
A-18.
If (1 + by) n = (1+ 8y + 24 y2 +....) where nN then the value of b and n are respectively(1) 4, 2 (2) 2, – 4 (3) 2, 4 (4) – 2, 4
A-19.
The coefficient of x 52 in the expansion
100
100
Cm (x – 3)100–m. 2m is :
m 0
(1) 100C47 A-20.
(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 + x 2) (1 + x + x 2 + x 3)...... (1 + x + x 2 +...... + x 100) when written in the ascending power of x then the highest exponent of x is (1) 5000 (2) 5030 (3) 5050 (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) 9C5. 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) none of these JEE(MAIN) BINOMIAL THEOREM - 17
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C-9.
50 The value of 0
50 50 + 1 1
100 (1) 50
100 (2) 51 10
C-10.
The value of
r .
r 1
n n
Cr
Cr 1
(1) 5 (2n – 9)
C-11.
50 50 +...........+ 2 49
50 n is, where nCr = 50 r 2
50 (3) 25
50 (4) 25
(3) 9 (n – 4)
(4) none of these
is equal to
(2) 10 n
10 10 Cr The value of the expression r 0 10 20 (1) 2 (2) 2
10 ( 1)K K 0
10
CK is : 2K
(4) 25
(3) 1
n
C-12.
C-13.
In the expansion of (1 +
1 1 , the term independent of x isx
(1) C20 + 2 C12 +.....+ (n + 1) Cn2
(2) (C0 + C1 +....+ Cn)2
(3) C20 + C12 +.......+ Cn2
(4) None of these
If (1 + x)n = C0 + C1x + C2x2 +...+Cn.xn then for n odd, C12 + C32 + C52 +.....+ Cn2 is equal to (2n)! (2n)! (1) 22n – 2 (2) 2n (3) (4) 2 2(n! ) (n! )2 n
C-14.
x)n
If an = (1)
r 0
n a 2 n
n
1 n
Cr
, the value of
n 2r
r 0
(2)
n
Cr
is :
1 a 4 n
(3) nan
(4) 0
Section (D) : Multinomial Theorem, Binomial Theorem for negative and fractional index D-1.
The coefficient of a5 b4 c 7 in the expansion of (bc + ca ab)8 is (1) 280 (2) 240 (3) 180
(4) 32
D-2.
If x < 1, then the co-efficient of x n in the expansion of (1 + x + x 2 + x 3 +.......)2 is (1) n (2) n 1 (3) n + 2 (4) n + 1
D-3
The coefficient of x4 in the expression (1 + 2x + 3x2 + 4x3 + ......up to )1/2 (where | x | < 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 x 6 in series e2x is (1)
4 45
(2)
3 45
JEE(MAIN) BINOMIAL THEOREM - 19
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4.
3 1 In the expansion of 4 4 6
20
(1) the number of irrational terms is 19 (3) the number of rational terms is 2 5.
6.
(2) middle term is irrational (4) All of these
If (1 + 2x + 3x 2)10 = a0 + a1x + a2x 2 +.... + a20x 20, then : (1) a1 = 20 (2) a2 = 210 (3) a4 = 8085
(4) All of these
(1 + x + x 2 + x 3)5 = a0 + a1x + a2x 2 +....................... + a15x 15, then a10 equals to : (1) 99 (2) 101 (3) 100 (4) 110 n
7.
1 3 In the expansion of x 2 , n N, if the sum of the coefficients of x 5 and x 10 is 0, then n is : x (1) 25
(2) 20
(3) 15
(4) None of these 10
8.
x 1 x 1 The coefficient of the term independent of x in the expansion of 2 1 1 3 x x3 1 x x2
(1) 70
(2) 112
(3) 105
The term in the expansion of (2x – 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) 805 (2) 301
(3) 503
(4) 803
79 + 97 is divisible by : (1) 7 (2) 24
(3) 64
(4) 72
12.
is :
(4) 210
9.
11.
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 of these
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) n1C2(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 ,T4 (2) T4 (3) T 5 , T 6 (4) T 6 , T 7
16.
Number of elements in set of value of r for which, 18Cr 2 + 2. 18Cr 1 + 18Cr 20C13 is satisfied : (1) 4 elements (2) 5 elements (3) 7 elements (4) 10 elements
17.
The number of values of ' r ' satisfying the equation, 39 C3r 1 39C
r2
(1) 1 18.
The sum
(1)
(2) 2
(3) 3
= 39 Cr 2 1 39 C 3r is : (4) 4
1 1 1 is equal to : ...... 1 ! ( n 1) ! 2 ! ( n 2) ! 1 ! ( n 1) !
1 (2n 1 1) n!
(2)
2 (2n 1) n!
(3)
2 n1 (2 1) n!
(4) none
JEE(MAIN) BINOMIAL THEOREM - 21
ANCE
PART - II : COMPREHENSION Comprehension # 1 Let P be a product given by P = (x + a1) (x + a2) ......... (x + an) n
and
Let S1 = a1 + a2 + ....... + an =
a , S i
i 1
2
=
a .a , S i
j
3
=
i j
a .a .a i
j
k
and so on,
i jk
then it can be shown that P = xn + S1 xn – 1 + S2 xn – 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 = C0 + C1 x + C2 x2 + C3x3 + ......+ C n x n . Various relations among binomial coefficients can be derived by putting
1 i 3 x = 1, – 1, i, where i 1, . 2 2 4.
The value of nC0 – nC2 + nC4 – nC6 + ....... must be (1) 2i (2) (1 – i)n – (1 + i)n
1 [(1 – i)n + (1 + i)n] 2
(3) 5.
(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
(3) 2 n
(2) 2n
2
(4) None of these
PART - I : AIEEE PROBLEMS (LAST 10 YEARS) 1.
If n
n
Cr 1 nCr 1 2 nCr equals
(1) 2.
Cr denotes the number of combinations of n things taken r at a time, then the expression
n2
Cr
(2)
n 2
[AIEEE 2003]
Cr 1
(3)
n 1
(4)
Cr
8
256
The number of integral terms in the expansion of
3 5
(1) 32
(3) 34
(2) 33
n 1
is :
Cr 1 [AIEEE 2003]
(4) 35. JEE(MAIN) BINOMIAL THEOREM - 23
ANCE
13.
The sum of the series 20C0 – 20C1 + 20C2 – (1) –20C10
14.
(2)
1 2
20
20
C3 + ..... + 20C10 is
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, the sum of 5th and 6th term is zero, then
(1)
n4 5
(2)
5 n4
(3)
6 n5
(4)
n5 6
n
15.
Statement-1 :
(r 1) nCr = (n + 2) 2n–1
[AIEEE 2008 (3, –1), 105]
r 0
n
Statement-2 :
(r + 1) nCr xr = (1 + x)n + nx (1 + x)n – 1 r 0
(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement-2 is True 10
16.
Let S1 =
10
j ( j – 1)
10
Cj , S2 =
j 1
10
j 10Cj and S3 =
j 1
j
2 10 C.
[AIEEE 2009 (4, –1), 144]
j
j 1
Statement -1 : S3 = 55 × 29 . Statement -2 : S1 = 90 × 28 and S2 = 10 × 28. (1) Statement -1 is true, Statement-2 is true ; Statement -2 is not a correct explanation for Statement -1. (2) Statement-1 is true, Statement-2 is false. (3) Statement -1 is false, Statement -2 is true. (4) Statement -1 is true, Statement -2 is true; Statement-2 is a correct explanation for 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 (1) an irrational number (3) an even positive integer
19.
2n
3 1
–
2n
3 1
[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
x 1 x 1 The term independent of x in expansion of 2 / 3 1/ 3 x 1 x x 1/ 2 x (1) 4
(2) 120
(3) 210
10
is : [AIEEE - 2013, (4, – ¼) 120 ] (4) 310
JEE(MAIN) BINOMIAL THEOREM - 25
ANCE
BOARD LEVEL SOLUTIONS
5. We have, ex = 1 +
Type (I) 1.
4
Putting x = 2, we get
4x
C0 x
3 4
1
º 4
+
4 C1( x 3 )3 + x 3
4
+
4 C 3 ( x 3 )1 + x
4
2.
4
4 C 2 ( x 3 )2 x
4 C4 (x ) x
x12 + 16 x8 + 96 x4 + 256 +
6
x x2 x3 x 4 + .... to 1! 2! 3 ! 4 !
6. loge(1 + 3x + 2x2) = loge[(1 + 2x) (1 + x)] = loge(1 + 2x) + loge(1 + x)
º b b C0 (ax) + 6C1 (ax)5 – x x
1
( 2 x )2 1 1 2 x – (2x )3 – (2x )4 ... = 2 3 4
2
b b + 6C2 (ax)4 – + 6C3 (ax)3 – x x 4
b b + C4 (ax) – + 6C5 ax – x x
3
1 2 1 3 1 4 + x – ( x ) ( x ) – ( x ) ... 2 3 4
5
2
b + C6 (ax) – x 6
2 22 23 2 4 25 26 27 +... 1! 2 ! 3 ! 4 ! 5 ! 6 ! 7 !
Ans.
x4
6
6
e2 = 1 +
e2 = 1 + 2 + 2 + 1·333 + 0.666 + 0.266 + 0.088 + 0.025 e2 = 7·378 e2 = 7·4 (correct to one decimal place)
4
3 0
256
2
= 3x –
5 2 17 4 x + 3x3 – x + ... 2 4
Type (II)
6
6
5
7.
C0 (1 + x)5 (–x2)0 + 5 C1 (1 + x)4 (–x2)1
= a6x6 – 6a5bx4 + 15a4b2x2 – 20a3b3 +
15a 2b 4 x2
4
3.
4
2
+
x4
0 4
=
x2 a2
C3 (1 + x)2 (–x2)3 + 5 C 4 (1 + x)1 (–x2)4 + 5 C5 (1 + x)0 (–x2)5
3
x a C3 a x
x a C4 a x
4x a a2 – +6–4 + 2 a x x
5
1
1 4
0
+
+
x6
x a C1 a x
2
x a + C2 a x 4
b6
+
3
x a + C0 a x 4
6ab 5 –
+ 5 C 2 (1 + x)3 (–x2)2
4
(1 + 5x + 10x2 + 10x3 + 5x4 + x5) + 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 10 –x + 5x9 – 5x8 – 10x7 + 15x6 + 11x5 – 15x4 – 10x3 + 5x2 + 5x + 1 Ans. º
Ans.
1 1 8. C0 (x + 2) + 3 C1 (x + 2)2 x x 3
1
3
2
4. e2x+3
( 2 x ) ( 2 x ) 2 ( 2 x )3 ... = e .e = e 1 1! 2! 3! 2x
3
3
Thus the coefficient of x2 in the expansion of e2x+3 is e3
1 1 + C 2 (x + 2) + 3 C3 (x + 2)0 x x 3
22 = 2e3 2!
3
1
1 [x3 + 8 + 12x + 6x2] + 3.[x2 + 4x + 4]. x
1 1 + 3(x + 2). 2 – 3 x x
JEE(MAIN) BINOMIAL THEOREM - 27
ANCE
16. As Tr+1 = n Cr x nr y r in (x + y)n
7 Now consider x x
7
1 1 7 = C 0 + C1 2 2
17
7
[on comparing n = 17, r = 10, x = x , y =
T11 =
17
C10 (x)17–10
T11 = 710
17
7 x
7 ] x
=
C10 x 3
17
C10 x 7 .
( 7 )10 x10
Ans.
100
C1 (100)1
C2 (100)2 +....... +
1 C2 2
5
1+
100
100
1 = 2[ C1 2
C100 (100)100
C2 +....... 10196]
C2 +..... + 10196 is a natural number by the
virtue of its being the binomial coefficients. = 104 N (101)100 – 1 is divisible by 10,000.
1 + C5 2
1995
C0 (7)
+
1995
1994
C1 (7)
= 1
(10)
+ ....
1995
C1995 (10)1995 +
+.....
1995
C1995 (10)1995 – 71995
1995
C0 + 10
Now
1995
C0 +
1995
2
+
1 + 10N [ 1995
C1995 (10)1994 +
n
...(ii)
1995
6
1 7 4x 1 + C3 2 2
4
3
4x 1 2
5
7
º 4x 1 + 7 C 7 1 4 x 1 ] 2 2 2
27
(4x + 1)2 +
( 4 x 1)3 ] 27
7 7 4 x 1 [ C1 + C3 (4x + 1)
+ 7 C5 (4x + 1)2 + (4x + 1)3]
C1995 (10)
1994
7 7 1 4 x 1 1 4 x 1 2 2 4 x 1
]
1
[ 7 C1 + 7 C3 (4x + 1) + 7 C5 (4x + 1)2 + (4x + 1)3] 26 It is a polynomial of degree 3. Ans. =
C1 (7)1994 +.......
1995
C1 +........
1995
C1995 (10)1994]
7
1 2
7
4 x 1 2
1
= N(natural number as it is the sum of binomial coefficients) Units place is 1 Ans.
19. Consider
26
C1 +......... 1995
1
1
C1 (10)1
[ 1995 C1(7)1994 +........ 1995 C1995 (10)1994 1995
4x 1 + 2
1 1 7 7 = 2 4 x 1 [ C1 . 7 + C 3 . 7 . (4x + 1) 2 2 7 + C5 .
18. Consider 171995 + 111995 – 71995 (7 + 10)1995 + (1 + 10)1995 – 71995 1995
6
2
7
100
4x 1 ...(i) C7 2
4 x 1 +...+ 7 C 7 2
7
Now (1+ 100)100 – 1 = 1 + 104 + 100 C2 104 +...10200 – 1 = 104 [1 +
7
(i) – (ii)
Using Binomial theorem
+
+.....+
7
7
100
7
4x 1 2
1 4 x 1 Now 2 2
[(1 + x)n = n C 0 + n C1 x + n C 2 x 2 +.......+ n Cn x n ]
C0 +
2
5
1 1 7 7 = C 0 + C1 2 2
17. (101)100 = (1 + 100)100
100
1 C2 2
4x 1 + 2
7
10
Type (III)
(1 + 100)100 =
6
7
4 x 1 n n n [ (x + y) = C 0 x + 2
C1 xn–1 y1 + n C 2 xn–2 y2 +....... + n Cn yn]
20. Let = x = t6 t6 1 t6 1 t 4 t2 1 t6 t3
10
( t 2 1) ( t 4 t 2 1) ( t 3 1) ( t 3 1) t 4 t2 1 t 3 ( t 3 1)
t 5 t 3 t 3 1 t3
10
10
JEE(MAIN) BINOMIAL THEOREM - 29
ANCE
(2n 1)! (2n 1)! = (2n 1)! (n 1)! (–1)r –1 + (r 1)! (2n r )! (–1)2n–r
1 1 1 1 = 2 1 – – ......to – 1 2 3 4 5
(2n 1)! = (2n r )! (r 1)! [(–1)r–1 + (–1)2n–r]
= 2 loge 2 – 1 = loge4 – logee = loge e = R.H.S.
4
1r 1 (2n 1)! = (2n r )! (r 1)! 1 ( 1)r
x x2 x3 28. We have e = 1 + +..... to 1! 2! 3 ! x
(2n 1)! = (2n r )! (r 1)! [0] = 0 proved.
Put x = 1, we get
[ (1 + x)n = n C 0 + n C1x + ....... + n Cn x n ] n
1
n
2
n
1 1 1 e + e–1 = 2 1 ..... 2! 4 ! 6 !
n
= 1 7n nC 2 (7)2 nC3 (7)3 ... nCn (7)n 7n 1 2
n
3
n
n
1 1 1 1 + .... = (e + e–1) 2! 4 ! 6 ! 2
Hence 1
= C 2 (7) C3 (7) ...... Cn (7)
2 x 4 x6 x8 ...... 29. L.H.S. = x 2! 3 ! 4 !
= 7 2 [ nC 2 nC 3 7 ...... nCn 7n2 ] = 49[ n C 2 nC 3 7 ...... nCn 7n2 ] n
C 2 nC 3 .7 ...... nCn 7n2 = N
2 y4 y6 ...... – 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 ) 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 =
n 1
( y 2 ) 2 ( y 2 )3 2 – 1 y 2! 3! ...
C0 n1C1(8)1 n1C2 (8)2 ....... n1Cn1(8)n1 8n 9
= 1 + (n + 1) 8 +
n 1
2
C 2 (8) ....... 8
n1
2
= ex – ey
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
1 1 1 1 1 1 1 = 1 – – – – – – + ...... 2 2 3 3 4 4 5
1 1 1 1 1 1 1 1 – – – .......... =1– 2 2 3 3 4 4 5 5
2
Type (IV) 30. Let loge 10 = x n
Now
r n
(1)
Cr
r 0
1 r x
( 1)r nCr
r 0
r n
(1)
Cr
r 0
x + 1 nx
n
1 +
(1 xn)r
n
[ log am = m log a]
(1 xn)r
n
1 1 1 1 – – 27. L.H.S. = + ..... to 1.2 2.3 3.4 4.5
1 1 1 1 = 1 – 2 – – .....to 2 3 4 5
1 1 1 – +....... 1! 2! 3!
add both equation, we get
= C 0 C1(7) C 2 (7) ...... Cn (7) 7n 1
n
1 1 1 + .... 1! 2! 3!
Put x = – 1, we get e–1 = 1 –
25. Consider 23n – 7n – 1 = (8)n – 7n – 1 = (1 + 7)n – 7n – 1
n
e=1+
r n
(1)
Cr
r 0
rx (1 nx )r
r (1 nx )r n
(1) r 1
r
n r
n 1
[using n Cr =
r
Cr 1
n r
n 1
(1 nx )r 1
Cr 1 ]
JEE(MAIN) BINOMIAL THEOREM - 31
ANCE
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 bd x [( C 4 ) C 3 . C 5 ]
35. As Tr+1 = n Cr xn–r yr in (x + y)n & consider (x + a)n
n ! n! n! n! 2 x (n 3)! (n 3)!3 ! 3 ! (n 2)! 2! (n 4)! 4 ! n! n! n! n! 2 (n 4 )! (n 4)! 4 ! 4 ! (n 3 )! 3 ! . (n 5)! 5 !
1 1 n! n! 1 x 2! 3 ! (n 3)! (n 4)! (n 3)3 4(n 2) n! n! 1 1 1 2 (n 4 )! (n 5)! (n 4 )4 5(n 3) 3 ! 4 !
x 2 4 ! (n 5)! 4n 8 3n 9 5(n 3) (n 4 ) . 2 2 ! (n 3) ! 3.4 (n 2).(n 3) 5n 15 4n 16
x 2 4 ! (n 5)! (n 1) 4.5(n 4) . 2 2 ! ( n 3 ) ! 12 .( n 2 ) (n 1)
=
x2
2
T4 = n C 3 xn–3 a3 = 1080
...(iii)
n
n (n 1) a a . = 3 (n – 1). = 6 ...(iv) 2n x x
(n – 2)
iv v
r 1
Tr+1 = Cr x
n n
r
Tr+2 = n Cr 1 x r 1
a 9 = x 2
Now it is given that coefficients of Tr , Tr+1 and Tr+2 are 76, 95, 76 repsectively.
Cr 1 = 76
...(i)
n
Cr = 95
...(ii)
...(v)
6 4 n 1 = .2 = 9 3 n2
a a 3 3x =6 = a= x x 2 2 3x in (i) n C1 xn–1 a1 = 240 2
Put a =
n
C 3 x n 3 a 3 1080 3 . n2 . 2 = = C2 x 720 2 a
3n – 3 = 4n – 8 n=5 From (iv) 4.
n
C 2 x n2 a 2 . =3 n C1 x n1 a1
n (n 1) (n 2).2 a 3 = 6n (n 1) x 2
As Tr+1 = n Cr x r in [1 + x]n
n
...(ii)
iii ii
34. Let Tr , Tr+1 and Tr+2 be the three consecutive terms in the expansion of (1 + x)n
Tr = Cr 1 x
T3 = n C 2 xn–2 a2 = 720
=
(n 4 ) ! 20 (n 2)!
n
(given) ...(i)
ii i
2
T2 = n C1 xn–1 a1 = 240
3x = 240 2 x5 = 32 x = 2 5.x4.
3x =3 2 n = 5; a = 3; x = 2 Now a =
Cr 1 = 76
Now
ii = i
...(iii) n n
Cr
Cr 1
=
95 nr 1 = 76 r
76n – 76r + 76 = 95 r 76(n + 1) = 101 r ...(iv)
iii = ii
n
36.
(i) Sum =
t
= t1 + t2 + t3 + ..... to
n
n 1
Cr 1
n
Cr
76 nr = = 95 r 1
95n – 95r = 76r + 76 95n – 76 = 101r ...(v) From (iv) 95n – 76 = 76n + 76 19n = 152 n=8 Ans.
Ans.
=
1
(n 1) ! n 1
=
1 1 1 + ..... to 2! 3! 4!
1 1 1 = 1 1! 2 ! 3 ! ..... – 2 = e – 2 (ii) We have, tn =
Sum =
1 (n 2) ! 1
(n 2 ) ! n 1
JEE(MAIN) BINOMIAL THEOREM - 33
ANCE
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 2x 2 1 2x 2 1 The expression, is a polynomial of degree 2 2 2x 1 2x 1
(1) 5 3.
4.
(2) 6
(4) 8
Co-efficient of x5 in the expansion of (1 + x2)5 (1 + x)4 is : (1) 40 (2) 50 (3) 30
(4) 60
Co-efficient of x15 in (1 + x +x3 + x4)n is : 5
(1)
5 n
n
C15 3r Cr
r 0
5.
(3) 7
(2)
5 n
C 5r
(3)
r 0
n
(4)
r 0
If n is even natural and coefficient of xr in the expansion of (2) r (n 2) / 2
(1) r n / 2
3
C 3r
n
n
C 3 r C 5r
r 0
1 x n 1 x
is 2n, (|x| < 1), then –
(3) r (n 2) / 2
(4) r n
6.
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
7.
n r 1
r 1
p 0
n
r Cr Cp 2p is equal to
(1) 4n – 3n + 1 8.
n
(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 n 3n 1 (1) 1 2 2
9.
(2) 4n – 3n – 1
3 n (2) 2 n 2
(3) 2n + 5n 2n
(4) (–2)n.
If the sum of the coefficients 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.
11*.
12.
1 2 The term independent of x in the expansion of ( 1 + x + 2x2) 3 x 2 is 3x (1) 10 (2) 2 (3) 0 (4) 6 1000n for n N, then an is greatest, when n! (1) n = 997 (2) n = 998 (3) n = 999
Let an
n 2k 0
(1) nCk
n n – 2k 1 k 1
n 1 + 2k 2 k 1
(2) n+1Ck
n 2
n n 2 –...... + (– 1)k k k 2 (3) n–1Ck
(4) n = 1000 n k = 0
(4) n+2Ck JEE(MAIN) BINOMIAL THEOREM - 35
