Answer
$$=32x^5+80x^4y+80x^3y^2+40x^2y^3+10xy^4+y^5 $$
Work Step by Step
The binomial theorem states that:
$$(a+b)^n=\sum _{i=0}^n\binom{n}{i}a^{(n-i)}b^i$$
Thus:
$$ (2x+y)^5 \\ =\frac{5!}{0!(5-0)!}(2x)^5y^0+\frac{5!}{1!(5-1)!}(2x)^4y^1+\frac{5!}{2!(5-2)!}(2x)^3y^2+\frac{5!}{3!(5-3)!}(2x)^2y^3+\frac{5!}{4!(5-4)!}(2x)^1y^4+\frac{5!}{5!(5-5)!}(2x)^0y^5 \\ =32x^5+80x^4y+80x^3y^2+40x^2y^3+10xy^4+y^5 $$
