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.
 A: You need to be a bit more careful with your notation, since as you have defined the expressions, $f(n)$ is the entire sum, and $g(n)$ is the value of that entire sum; therefore, you should always have $$f(n) = g(n).$$
What you should be observing is that since $f(n)$ is composed of a finite sum of individual terms of the form $k \cdot 3^k$, the addition of another term of the form $(n+1)3^{n+1}$ to $f(n)$ should yield $g(n+1)$.  This is the induction hypothesis, and so it should be written $$f(n+1) = f(n) + (n+1)3^{n+1} \overset{\text{i.h.}}{=} g(n) + (n+1) 3^{n+1} = g(n+1).$$  The $\text{i.h.}$ over the equals sign is the step in the chain of equalities where the induction hypothesis is used. What you need to show is the rightmost equality in the chain.
A: You can calculate $$\frac{3^{n+2}(2n+1)-3-3^{n+1}(2n-1)-3}{4}=\frac{3^{n+1}(3(2n+1)-2n+1)}{4}=\frac{3^{n+1}(6n+3-2n+1)}{4}=\frac{3^{n+1}(6n+3-2n+1)}{4}=…$$
A: We have
$$\sum_{k=1}^{n+1} k\cdot 3^k=(n+1)\cdot 3^{n+1}+\sum_{k=1}^n k\cdot 3^k\stackrel{Ind. Hyp.}=(n+1)\cdot 3^{n+1}+\frac {3(3^n(2n-1)+1)} 4=$$
$$=\frac {4(n+1)\cdot 3^{n+1}+3^{n+1}(2n-1)+3} 4=\frac {3^{n+1}(6n+3)+3} 4=$$
$$=\frac {3(3^{n+1}(2(n+1)-1)+1)} 4=f(n+1)$$
A: It's better, to prove this equals in another way. 
$\sum_{k=1}^n k*3^k= \sum_{k=1}^n 3^k +  \sum_{k=2}^n 3^k +  \sum_{k=3}^n 3^k +... =  \sum_{i=1}^n \sum_{k=i}^n 3^k $
A: To do the inductive step, observe that:
$(n+1)3^{n+1}+\frac{3(3^n(2n-1)+1)}4=\frac{4(n+1)3^{n+1}+3(3^n(2n-1)+1)}4=\frac{(6n+3)3^{n+1}+3}4=\frac{3(2n+1)3^{n+1}+3}4=\frac{3(3^{n+1}(2(n+1)-1)+1)}4$...
