Induction. Am I missing something or is there a mistake in the question? $$\sum_{k=1}^n k*3^k=\frac {3(3^n(2n-1)+1)} 4 $$
So let f(n)= $\sum_{k=1}^n k*3^k $
and  g(n)=$\frac {3(3^n(2n-1)+1)} 4$
By induction hypothesis, $f(n+1) = f(n) + (n+1)3^{n+1} \overset{\text{i.h.}}{=} g(n) + (n+1) 3^{n+1} = g(n+1).$
$$\frac{3(3^{n+1}(2n+1)+1)}4=(n+1)(3^{n+1})+\frac{3(3^n(2n-1)+1))}4 $$
I am stuck afterward, please help, thanks.

We have
$$\sum_{k=1}^{n+1} k\cdot 3^k=(n+1)\cdot 3^{n+1}+\sum_{k=1}^n k\cdot 3^k\stackrel{Ind. Hyp.}=(n+1)\cdot 3^{n+1}+\frac {3(3^n(2n-1)+1)} 4=$$
$$=\frac {4(n+1)\cdot 3^{n+1}+3^{n+1}(2n-1)+3} 4=\frac {3^{n+1}(6n+3)+3} 4=$$
$$=\frac {3(3^{n+1}(2(n+1)-1)+1)} 4=f(n+1)$$
