Answer
$x^5+5x^4+10x^3+10x^2+5x+1$
Work Step by Step
According to the binomial theorem, we can expand the algebraic expression in the form of:
$(a+b)^n=\binom{n}{0}b^0a^n+\binom{n}{1} b^1a^{n-1}++\binom{n}{n-i} b^{n-i}a^{i}++\binom{n}{n-1}b^{n-1}a^1+\binom{n}{n}b^nx^0$
Now, we will expand the given expression by replacing $b=1$ and $n=5$.
$(x+1)^5=\binom{5}{0}x^51^0+\binom{5}{1} x^41^1+\binom{5}{2} x^{3}1^{2}+\binom{5}{3}x^{2}1^3+\binom{5}{4}x^11^4+\binom{5}{5}x^01^5 \\=x^5+5x^4+10x^3+10x^2+5x+1$
