Question

Question #00978

Answer 1

See below.

Explanation:

`(1+ax)^n=1+nax+(n(n-1))/(2!)(ax)^2+(n(n-1)(n-2))/(3!)(ax)^3+cdots`

The coefficient of `x` is `na` and also

`a^3(n(n-1)(n-2))/(3!)=6a^2(n(n-1))/(2!)`

so the conditions are

`{(n a=12),(a(n-2)/(3!)=6/(2!)):}`

solving for `a,n` we have

`n = -4` and `a = -3` or

`1/(1-3x)^4`

and the expansion is valid for `x ne 1/3`