Answer
$(3x+1)^4=\binom{4}{0}(3x)^41^0+\binom{4}{1}(3x)^{3}1^1+\binom{4}{2}(3x)^{2}1^2+\binom{4}{3}(3x)^{1}1^3+\binom{4}{4}(3x)^{0}1^{4}=81x^4+108x^3+54x^2+12x+1$
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, x becomes 3x, $y=1; n=4$
$(3x+1)^4=\binom{4}{0}(3x)^41^0+\binom{4}{1}(3x)^{3}1^1+\binom{4}{2}(3x)^{2}1^2+\binom{4}{3}(3x)^{1}1^3+\binom{4}{4}(3x)^{0}1^{4}=81x^4+108x^3+54x^2+12x+1$
