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12th Grade > Mathematics

BINOMIAL THEOREM MCQs

Binomial Theorem

Total Questions : 74 | Page 1 of 8 pages
Question 1. If the third term in the binomial expansion of (1+x)m is -18x2, then the rational value of m is
  1.    2
  2.    12
  3.    3
  4.    4
 Discuss Question
Answer: Option B. -> 12
:
B
mc2=18m(m1)2=184m24m+1=0(2m1)2=0m=12
Question 2. If a1,a2,a3,a4 are the coefficients of any four consecutive terms in the expansion of (1+x)n, then a1a1+a2+a3a3+a4= 
  1.    a2a2+a3
  2.    12 a2a2+a3
  3.    2a2a2+a3
  4.    2a3a2+a3
 Discuss Question
Answer: Option C. -> 2a2a2+a3
:
C
Let a1,a2,a3,a4 be respectively the coefficients of (r+1)th,(r+2)th,(r+3)thand(r+4)th terms in the expansion of (1+x)n.
Then a1=nCr,a2=nCr+1,a3=nCr+2,a4=nCr+3
Now,
a1a1+a2+a3a3+a4=nCrnCr+nCr+1+nCr+2nCr+2+nCr+3=nCrn+1Cr+1+nCr+2n+1Cr+3=nCrn+1r+1nCr+nCrn+1r+3nCr+2=r+1n+1+r+3n+3=2(r+2)n+1
Also, solving the R.H.S, we get
2a2a2+a3=2nCr+1nCr+1+nCr+2=2nCr+1n+1Cr+2=2(r+2)n+1
Question 3. The value of (2+1)6 + (21)6 will be
  1.    -198
  2.    198
  3.    99
  4.    -99
 Discuss Question
Answer: Option B. -> 198
:
B
(x+a)n + (xa)n = 2[xn+nC2xn2a2+nC4xn4a4+nC6xn6a6+.......]
Here, n = 6, x =2, a = 1;
6C2 = 15,6C4 = 15,6C6 = 1
∴ (2+1)6 + (21)6 = 2[(2)6+15.(2)4.1+15(2)2.1+1.1]
=2[8+15×4+15×2+1]=198
Question 4. If x4 occurs in the rth term in the expansion of (x4+1x3)16, then r =
  1.    7
  2.    8
  3.    9
  4.    10
 Discuss Question
Answer: Option C. -> 9
:
C
Tr =16Cr1 (x4)16r(1x3)r1 =16Cr1x677r
677r=4r=9
Question 5. The first 3 terms in the expansion of (1+ax)n (n ≠ 0) are 1, 6x and 16x2. Then the value of a and n are respectively
  1.    2 and 9
  2.    3 and 2
  3.    23 and 9
  4.    32 and 6
 Discuss Question
Answer: Option C. -> 23 and 9
:
C
nc1.ax=6x;nc2(ax)2=16x2an=6;n(n1)2a2=16a=23n=9
Question 6. In (32+133)n if the ratio of 7th term from the beginning to the 7th term from the end is  16, then n =
  1.    7
  2.    8
  3.    9
  4.    None of these
 Discuss Question
Answer: Option C. -> 9
:
C
16=nC6(213)n6(313)6nCn6(213)6(313)n6
61=613(n12)
n12=3
n=9
Question 7. Find the value of 
(183+73+318725)36+62432+15814+20278+15916+6332+64
 
  1.    1
  2.    5
  3.    25
  4.    100
 Discuss Question
Answer: Option A. -> 1
:
A
The numerator is of the form
a3 + b3+3ab(a+b)=(a+b)3
∴ N' = (18+7)3 = 253
∴ D' = 36+6C13521+6C23422+6C33323+6C43224+6C53125+6C626
This is clearly the expansion of (3+2)6=56=(25)3
ND=(25)3(25)3=1
Question 8. The number of terms which are free from radical signs in the expansion of (y15+x110)55 is
  1.    5
  2.    6
  3.    7
  4.    None of these
 Discuss Question
Answer: Option B. -> 6
:
B
In the expansion of(y15+x110)55, the general
term is
Tr+1 =55Cr(y15)55r(x110)r =55Cry11r5xr10.
This Tr+1 will be independent of radicals if the exponents r5 and r10 are integers, for
0≤ r≤ 55 which is possible only when
r = 0, 10, 20, 30, 40, 50.
∴ There are six terms viz. T1,T11,T21,T31,T41,T51
which are independent of radicals.
Question 9. The value of (2+1)6 + (21)6 will be
  1.    -198
  2.    198
  3.    99
  4.    -99
 Discuss Question
Answer: Option B. -> 198
:
B
(x+a)n + (xa)n = 2[xn+nC2xn2a2+nC4xn4a4+nC6xn6a6+.......]
Here, n = 6, x =2, a = 1;
6C2 = 15,6C4 = 15,6C6 = 1
∴ (2+1)6 + (21)6 = 2[(2)6+15.(2)4.1+15(2)2.1+1.1]
=2[8+15×4+15×2+1]=198
Question 10. If the coefficient of (2r+4)th and (r2)th terms in the  expansion of (1+x)18 are equal, then r =
  1.    12
  2.    10
  3.    8
  4.    6
 Discuss Question
Answer: Option D. -> 6
:
D
18C2r+3 =18Cr3 ⇒ 2r + 3 + r - 3 = 18 ⇒ r = 6

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