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?

1 Answer
Nov 26, 2017

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.