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
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 = n−r−1r+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.