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Question

The coefficients of three successive terms in the expansion of (1+x)n are 165, 330 and 462 respectively, then the value of n will be


A

11

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B

10

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C

12

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D

8

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Solution

The correct option is A

11


Let the coefficient of three consecutive terms

i.e. (r+1)th, (r+2)th, (r+3)th in expansion of

(r+1)n are 165, 330 and 462 respectively then,

Coefficient of (r+1)th term = nCr = 165

Coefficient of (r+2)th term = nCr+1 = 330 and

Coefficient of (r+3)th term = nCr+2 = 462

nCr+1nCr = 2

or n - r = 2(r + 1) or r = 13(n - 2)

and nCr+2nCr+1 = nr1r+2 = 231165

or 165(n - r - 1) = 231(r + 2) or 165n - 627 = 396r

or 165n - 627 = 396 × 13 × (n - 2)

or 165n - 627 = 132(n - 2) or n = 11.


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