<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] Notice that
$$3^{n+1}(n+1) = 3 \cdot 3^n (n+1) = \frac{3}{4} \cdot 3^n \cdot 4(n+1)$$
Because of the 3/4, you can move this term into the parentheses and combine it with the $3^n(2n-1)+1$ part. And because of the $3^n$, you can move the $4(n+1)$ into the parentheses and combine with the $2n-1$ part, to get
$$ \frac{3}{4} \left( 3^n (2n-1) + 1 \right) + 3^{n+1}(n+1) = \frac{3}{4} \left( 3^n \left(6n+3 \right) + 1 \right) $$
Notice that $6n+3 = 3(2n+1)$, so in the parentheses you get $3^{n+1}(2n+1)+1$.
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