Question

expand (2a+b^2)^5

Answer 1

You need the binomial theorem:
\[(a+b)^n = \sum_{k=0}^n a^{n-k}b^k \frac{n!}{k!(n-k)!}\]
In this case you have n = 5:
\[a^5b^0\frac{5!}{0!(5-0)!}+a^4b^1 \frac{5!}{1!(5-1)!}+…\]
With:
\[0! = 1! = 1\]