Answer
$y=1; n=5$ $(x+1)^5=\binom{5}{0}x^51^0+\binom{5}{1}x^{4}1^1+\binom{5}{2}x^{3}1^2+\binom{5}{3}x^{2}1^3+\binom{5}{4}x^{1}1^{4}+\binom{5}{5}x^01^5=x^5+5x^4+10x^3+10x^2+5x+1$
Work Step by Step
The binomial theorem expands an algebraic expression in the form of: $(x+y)^n=\binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+\binom{n}{2}x^{n-2}y^2+...+\binom{n}{i}x^{n-i}y^i+...+\binom{n}{n-1}x^{1}y^{n-1}+\binom{n}{n}x^0y^n=\sum_{i=0}^{n}\binom{n}{i}x^{n-i}y^i$
Here:
$y=1; n=5$ $(x+1)^5=\binom{5}{0}x^51^0+\binom{5}{1}x^{4}1^1+\binom{5}{2}x^{3}1^2+\binom{5}{3}x^{2}1^3+\binom{5}{4}x^{1}1^{4}+\binom{5}{5}x^01^5=x^5+5x^4+10x^3+10x^2+5x+1$
