BINOMIAL THEOREM-JEE(MAIN)

ANCE

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

nr 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

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

n

(1 + x) =



n

Cr x r

r 0

JEE(MAIN) BINOMIAL THEOREM - 1

ANCE

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 

96

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

   


ANCE

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 r4

5.


ANCE

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

r0

(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.


ANCE

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 n1   C2  .....  Cn   = C 0 x  C1  2 3 n  1  n  1  1  1 C1 C 2 C   1   .....  ( 1)n1 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

r0



2.r . (nCr)2 +

r 0

(

n

Cr )2

r0

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



n1

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

ANCE

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

r0

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

73 (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


ANCE

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  21   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


ANCE

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          .... mn 3mn 5mn  

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

ANCE

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 nN 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

ANCE

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


ANCE

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) 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 ,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 n1 (2  1) n!

(4) none


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 jk

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

n2

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)

n4 5

(2)

5 n4

(3)

6 n5

(4)

n5 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


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


ANCE

16. As Tr+1 = n Cr x nr 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


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 7n2 ] = 49[ n C 2  nC 3 7  ......  nCn 7n2 ] n

C 2  nC 3 .7  ......  nCn 7n2 = 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  n1C1(8)1  n1C2 (8)2  .......  n1Cn1(8)n1  8n  9

= 1 + (n + 1) 8 +

n 1

2

C 2 (8)  .......  8

n1

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 ]


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 n2

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 . n2 . 2 = = C2 x 720 2 a

3n – 3 = 4n – 8 n=5 From (iv) 4.

n

C 2 x n2 a 2 . =3 n C1 x n1 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 nr 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 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.

=

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


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

BINOMIAL THEOREM-JEE(MAIN)

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