What is missing in this mathematical induction?

Question

Any idea where the missing 3^k+2 comes from? (sorry for the format, this thing didn't allow me to post images)

Answer 1

The last step is simply $$(k+1)(3^{k+1})+\frac{(2k-1)(3^{k+1})+3}{4}$$ $$\frac{4(k+1)(3^{k+1})+{(2k-1)(3^{k+1})+3}}{4}$$ $$\frac{(3^{k+1})(4k+4+2k-1)+3}{4}$$ $$\frac{(3^{k+2})(2k+1)+3}{4}$$ They have simply taken the $3$ out $$3\left(\frac{(3^{k+1})(2k-1)+1}{4}\right)$$

Answer 2

Here's the inductive step:

We have that:

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

So now consider:

$1\cdot 3 + 2\cdot 3^2 +... + n\cdot 3^n + (n+1)\cdot3^{n+1} = \frac{(2n-1)3^{n+1}+3}{4} + (n+1)3^{n+1}$

$= \frac{(2n-1)3^{n+1}+3+4(n+1)3^{n+1}}{4}$ =$ \frac{6n\cdot 3^{n+1}+3\cdot 3^{n+1} +3}{4} $= $\frac{3^{n+2}(2n+1)+3}{4}$ as required.

Hence true $\forall n\in \mathbb{Z^{+}}$