Question

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

Answer 1

80

Explanation:

Binomial theorem:

`(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k)y^k`

`(x+2)^5 = sum_(k=0)^5((5),(k)) x^(5-k)2^k`

Looking for `x^2` so look at the `k=3` term:

`((5),(3))x^2*2^3 = 8*(5!)/(3!2!)x^2 = 80x^2`