Binomil Therom

BINOMIAL THEOREM & MATHEMATICAL INDUCTION 1


EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS
Binomial Co-efficient (c) 7 (d) none of these
1. Which of the following is not true ? Ans. (b)
n n n n n+1
(a) Cr = Cn–r (b) Cr + Cr–1 = Cr Sol. Given :
(c) r. n Cr  n. n 1Cr 1 (d) nCr + nCr–1 = nCr+1  n  2 !  210  n  1 !
Ans. (d)
On expanding,
Sol. Since
 n  2   n  1 n   n  1!  210  n  1!
n
Cr  n Cr 1  n  1Cr
Hence (d) is the incorrect option.
 n  2   n  1  n   7  6  5
2. If n is a positive integer; then nC0 is equal to n = 5.
(a) n (b) 0 1 1 x
5. If   , then x is equal to
(c) 1 (d) none of these 8 9 10
Ans. (c)
(a) 100 (b) 90
Sol. We know that,
(c) 170 (d) none of these
n n! Ans. (a)
C0 
(n  0)!0! and 0! = 1
1 1 x
Sol. Given :  
8! 9! 10!
 n C0  1
Multiplying by 10! on both sides,
3. If 2nC3 : nC2 : : 44 : 1, then the value of n is
(a) 17 (b) 6 10! 10!
 x
(c) 11 (d) none of these 8! 9!
Ans. (a)
10.9.8...1 10.9...1
Sol. Given :  x
8.7...1 9.8...1
2n! 2!(n  2)! 90 + 10 = x
,  44
3! (2n  3)! n! x = 100
On expanding, 6. If nC3 = nC2, then n is equal to
(a) 2 (b) 3
2n(2n  1) (2n  2) (2n  3)...1 2.(n  1)!
,  44 (c) 5 (d) none of these
3.2 (2n  3)! n(n  1) (n  2)...1
Ans. (c)
4(2n  1) n
.1  44 Sol. C3  n C2  n  3  2
3
OR n = 5.
2n – 1 = 33
7. If nC8 = nC6 , then nC2
2n = 34
(a) 21 (b) 20
n = 17
(c) 91 (d) 28
4. If n  2  210 n  1, then the value of n is equal to Ans. (c)
(a) 6 (b)5

, 2 BINOMIAL THEOREM & MATHEMATICAL INDUCTION

n Sol. Given nC4, nC5, nC6 are in A.P.
Sol. Given C8  n C6
n n n
 n  8  6 OR n  14  2       
5 4 6
n 14! 14  13
Now C2  14C2    91
2! 12! 2 n! n! n!
2  
(n  5)!5! (n  4)!4! (n  6)!6!
8. If 32C2n–1 = 32Cn–3 , then n =
(a) 10 (b) 9 2 1 1
  
(c) 12 (d) 11 (n  5)5 (n  4) (n  5) 6.5
Ans. (c)
Taking LCM and on simplying we get,
32 32
Sol. C2n1  Cn3  6 (2n – 13) = (n – 4) (n – 5)
32 = 2n – 1 + n – 3  12n – 78 = n2 – 9n + 20
3n = 36  n2 – 21 + 98 = 0
n = 12.  (n – 14) (n – 7) = 0
9. If n+1C4 = 9 nC2 , then n = Thus n = 14 or n = 7
(a) 10 (b) 9 Binomial Theorem
(c) 12 (d) 11 12. The expansion (x + a)n = nC0 xn + nC1 xn–1 a1 + ..... + nCn an is valid
Ans. (d) when n is
n+1
Sol. C4 = 9 . nC2 (a) an integer (b) a natural number
(c) a rational number (d) none of these
n 1 n n 1 9
  9.  
4 n 3 2 n  2 12  1 n  2 Ans. (b)
n
Sol. Cr is defined for n  N
 (n + 1) (n – 2) = 12 × 9
 n = 11 13. ( x  x 3  1) 5  ( x  x 3  1 ) 5 is a polynomial of degree
10. If nCr–1 = 36, nCr = 84 and nCr+1 = 126, then r = (a) 5 (b) 6
(a) 1 (b) 2 (c) 7 (d) 8
(c) 3 (d) 4 Ans. (c)
Ans. (c)
Sol. ( x  x 3  1) 5  ( x  x 3  1 ) 5
n
n
Cr 84 r nr 7 n  r 1 7  2 4
Sol. n
Cr 1
 
36 n
 
3 r

3
...(i)  2  5 C0 x 5  5 C 2 x 3

 
x 3  1  5 C4 x.  x3  1 


r 1 n  r 1
2
n
 2  5C0 x 5  5 C2 x 3 x 3  1  5C4 x x 3  1
    

Cr 1 126 nr 3  
n
   ...(ii)
Cr 84 r 1 2
 degree of the polynomial = 7
From (i) and (ii) n = 9, r = 3 14. The total number of terms in the expansion of
11. If nC4, nC5 and nC6 are in A.P., then possible value of n is (x + a)100 + (x – a)100 after simplification is

(a) 6 (b) 12 (a) 202 (b) 51

(c) 14 (d) 21 (c) 50 (d) None

Ans. (c) Ans. (b)