Question

How do you find the coefficient of `x^2` in the expansion of `(x+3)^5`?

Answer 1

Coefficient of `x^2` is `270`

Explanation:

Binomial expansion of `(x+a)^n` is

`Sigma_(r=0)^(r=n)((n),(r))x^(n-r)a^r`, where `((n),(r))=(n!)/((n-r)!r!)`

Here `n=5` and as we seek coefficient of `x^2`,

`n-r` i.e. `5-r=2` or `r=3` and `a=3`

Hence coefficient of `x^2` is

`((n),(r))a^r=((5),(3))3^3=(5!)/(2!3!)27=(5xx4xx3xx2xx1)/(2xx1xx3xx2xx1)xx27=270`