Question

In the binomial expansion of `(1+ax)^n`, where `a` and `n` are constants, the coefficient of `x` is 15. The coefficient of `x^2` and of `x^3` are equal. What is the value of `a` and of `n`?

Answer 1

`a=6, n=2.5`

Explanation:

The formula for the binomial expansion of `(1+ax)^n` is:
`1+n(ax)+(n*(n-1))/(2!) (ax)^2...(n(n-1)...(n-r+1))/(r!) (ax)^r`

Therefore the `x^1` coefficient is `an=15`
If the `x^2` and `x^3` coefficients are equal, this must mean that:
`(n(n-1))/(2!)(a)^2=(n(n-1)(n-2))/(3!)(a)^3`

Taking out factors of `(n(n-1))/2 a^2` gives: `1=(n-2)/3a`

which expands to: `3=an-2a`

We know that `an=15`, so `3=15-2a` which rearranges to `2a=12`, so `a=6`.

`an=15` so `n=15/a=15/6=2.5`

You can check the answer by putting it back into the original formula.