The first three terms in the expansion of `(1+a x)^n(n!=0)` are `1,6x and 16 x^2dot` Then find the value of `a and ndot`

Answer

`a=2//3, n = 9`

Explanation

`T_(1) = .^(n)C_(0)=1`
`T_(2) = .^(n)C_(1) ax = 6x " "(1)`
`T_(3) = .^(n)C_(2) (ax)^(2) = 16x^(2) " "(2)`
From (2),
`na = 6 " " (3)`
From (3),
`(n(n-1))/(2)a^(3) = 16 " "(4)`
Eliminating a from (3) and (4),
`(n-1)/(2n) = 4/9` or `n = 9`
From (3),
`a = 2//3`