<s>[INST] Prove by induction that $1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$ I'm trying to prove this using induction
$1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$
So far I have:

*

*Base case: true

*Induction step:
$\dfrac{3}{4}(3^n(2n-1)+1)+(n+1)*3^{n+1}=\dfrac{3}{4}(3^{n+1}(2(n+1)-1)+1)$
here is where I get stuck on simplification of both left and right side so that they are equal
$\dfrac{3}{4}(3^n(2n-1)+1)+3^{n+1}+3^{n+1}n=\dfrac{3}{4}(3^{n+1}(2(n+1)-1)+1)$
Thanks for your help!
 [/INST] In the induction step you want to prove the increment from $n$ to $n+1$ on both sides are the same, namely,
$$(n+1)3^{n+1} = \dfrac{3}{4}(3^{n+1}(2n+1)-3^n(2n-1))\tag1$$
By looking at the increment you got rid of the annoying constant $1$ from the RHS. Then you notice immediately that you can divide $3^{n+1}$ from both sides, and $(1)$ becomes
$$n+1 = \dfrac{1}{4}(3(2n+1)-(2n-1))=\frac 14 (6n+3-2n+1)=\frac 14 (4n+4)=n+1$$
then you are done.
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