Induction. Am I missing something or is there a mistake in the question?

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.

Answer 1

You can calculate $$\frac{3^{n+2}(2n+1)-3-3^{n+1}(2n-1)-3}{4}=\frac{3^{n+1}(3(2n+1)-2n+1)}{4}=\frac{3^{n+1}(6n+3-2n+1)}{4}=\frac{3^{n+1}(6n+3-2n+1)}{4}=…$$

Answer 2

We have

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

$$=\frac {3(3^{n+1}(2(n+1)-1)+1)} 4=f(n+1)$$

Answer 3

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.

Answer 4

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 $

Answer 5

To do the inductive step, observe that: $(n+1)3^{n+1}+\frac{3(3^n(2n-1)+1)}4=\frac{4(n+1)3^{n+1}+3(3^n(2n-1)+1)}4=\frac{(6n+3)3^{n+1}+3}4=\frac{3(2n+1)3^{n+1}+3}4=\frac{3(3^{n+1}(2(n+1)-1)+1)}4$...