Answer: [tex]x^{10}+10x^8y+40x^6y^2+80x^4y^3+80x^2y^4+32y^5[/tex]
Step-by-step explanation:
[tex]\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i[/tex]
[tex]a=x^2,\:\:b=2y[/tex]
[tex]=\sum _{i=0}^5\binom{5}{i}\left(x^2\right)^{\left(5-i\right)}\left(2y\right)^i[/tex][tex]=\frac{5!}{0!\left(5-0\right)!}\left(x^2\right)^5\left(2y\right)^0+\frac{5!}{1!\left(5-1\right)!}\left(x^2\right)^4\left(2y\right)^1+\frac{5!}{2!\left(5-2\right)!}\left(x^2\right)^3\left(2y\right)^2+\frac{5!}{3!\left(5-3\right)!}\left(x^2\right)^2\left(2y\right)^3+\frac{5!}{4!\left(5-4\right)!}\left(x^2\right)^1\left(2y\right)^4+\frac{5!}{5!\left(5-5\right)!}\left(x^2\right)^0\left(2y\right)^5[/tex]
simplify all = [tex]x^{10}+10x^8y+40x^6y^2+80x^4y^3+80x^2y^4+32y^5[/tex]
