Binomial Therom

BINOMIAL THEOREM & MATHEMATICAL INDUCTION 11


EXERCISE - 2 : PREVIOUS YEAR JEE MAINS QUESTIONS
1. The number of integral terms in the expansion of Ans. (b)
256 Sol. (1 + x) (1 – x)n = (1 – x)n + x(1 – x)n
 3 8 5  is (2003)
 Coefficient of xn is = (–1)n + (–1)n–1 nC1
(a) 32 (b) 33 = (–1)n [1 – n]
(c) 34 (d) 35 4. The coefficient of the middle term in the binomial expansion
Ans. (b) in powers of x of (1 + x)4 and of (1 – x)6 is the same, if 
equals (2004)
r
256  r
Sol. Tr 1  256
Cr   3 58
(a) 
5
(b)
10
3 3
256  r r
For integral terms , are both positive integers 3 3
2 8 (c)  (d)
10 5
 r = 0, 8, 16, ...... 256 Ans. (c)
 256 = 0 + (n – 1)8 (using tn = a + (n – 1)d) Sol. Coefficient of middle term in (1  x)6
n = 32 + 1  n = 33 3
4
C2  2  6 C3   
n n
1 r t
2. If s n   n and t n   n then n is equal to 3
r 0 Cr r 0 C r s n 
10
(2004)
11
 1 
n n 5. If the coefficient of x7 in ax 2   equals the coefficient
(a) (b) 1  bx 
2 2
of x–7 in then a and b satisfy the relation
(2005)
2n  1
(c) n – 1 (d)
2 a
(a) ab = 1 (b) 1
Ans. (a) b

n n n (c) a + b = 1 (d) a – b = 1
n  (n  r) 1 nr
Sol. tn   n
 tn  n n
 n Ans. (a)
r 0 Cr r 0 Cr r 0 Cn  r
Sol.
n n
(Remark:using Cr  Cn  r ) 11 r
 1  11 r  1  a11 r 22  3r
n
Tr 1 of  ax 2    11Cr ax 2
 bx 
   
 bx 
 11
C r
br
x
1 r
tn  n n
 n (replacing n – r by r)
Cr r 0 Cr 11 r
 1  11  r  1  r a11 r 11 3 r
Tr 1 of  ax  2   11C r  ax    2   1
11
Cr x
tn = nsn – tn  bx   bx  br

tn n  2 1 
11
a 6  22  3r  7 
s 2 11
 Coefficient of x7 in  ax    C5 5  
n  bx  b  r  5 
3. The coefficient of xn in expansion of (1 + x) (1 –x)n is
 1  11 a5
11
11  3r  7 
(2004) and coefficient of x–7 in  ax  2   C6 6  
 bx  b  r  6 
(a) (n – 1) (b) (–1)n (1 – n)
(c) (–1)n–1 (n–1)2 (d) (–1)n–1 n

, 12 BINOMIAL THEOREM & MATHEMATICAL INDUCTION

6
a6 a5 50 56  r
Now
11
C5  11C6 C4   C3  56 C 4
5 6
b b r 1

 ab = 1. 8. For natural numbers m, n if (1–y)m (1 + y)n =1 + a1y + a2y2 +...
6. If the coefficients of rth, (r + 1) th and (r + 2)th terms in the and a1 = a2 = 10, then (m, n) is (2006)
binomial expansion of (1 + y)m are in AP, then m and r (a) (35, 20) (b) (45, 35)
satisfy the equation (2005) (c) (35, 45) (d) (20, 45)
(a) m2 – m (4r – 1) + 4r2 + 2 = 0 Ans. (c)
(b) m2 – m (4r + 1) + 4r2 – 2 = 0 Sol. Using binomial expansion:
(c) m2 – m (4r + 1) + 4r2 + 2 = 0
(d) m2 – m (4r – 1) + 4r2 – 2 = 0 m n  m  m  1 2 
1  y  1  y    1  my  y  ....... 
 2! 
Ans. (b)

 n  n  1 2 
Sol. Given mCr–1, mCr, mCr+1 are in A.P 1  ny  y  ..... 
 2! 
 2. m Cr  m Cr 1  m Cr 1
 m  m  1 n  n  1 
m
Cr 1 m
Cr 1  1  y n  m  y2    mn  + ........
 2.    2 2 
m m
Cr Cr
so n – m = 10 – (1)  a1  10
r mr
2  
m  r 1 r 1 m 2  m  n 2  n  2mn
 10  a2  10
2 2
2
 m  m (4r  1)  4r  2  0
2

6
m  n   m  n   20 –(2)
50 56  r
7. The value of C4   C3 is (2005) Solving (1) &(2) given (35, 45).
r 1
9. The sum of the series
(a) 56C4 (b) 56C3
20
C0 – 20C1 + 20C2 – 20C3 + ... + 20C10 is (2007)
(c) 55C3 (d) 55C4
1 20
Ans. (a) (a) – 20C10 (b) C10
2
6 (c) 0 (d) 20C10
Sol. 50
C 4   56 r C3
r 1 Ans. (b)

 50 C4   55 C3  54 C3  53 C3  52 C3 51 C3  520 C3  Sol. (1 + x)20 = 20C0 + 20C1x + 20C2 x2 + ... 20C20 x20
put x = –1
= (50C4 + 50C3) + 51C3 + 52C3 + 53C3 + 54C3 + 55C3
0 = 20C0 – 20C1 + 20C2 .... + 20C20

 using n
Cr 1  n Cr  n 1Cr 1  0 = 2(20C0 – 20C1 + ... – 20C9) + 20C10

1 20
= (51C4 + 51C3) + 52C3 + 53C3 + 54C3 + 55C3  20 C0  20 C1  .......  20 C9  C10
2
in the same manner
20
1 20
 C0 – 20C1 + ... + 20C10 = . C10
2