Prove by Induction $1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ n(3n-1) = n^2(n+1)$ Prove by induction that the following equality holds true for all n that's an element of a natural number.
$$1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ n(3n-1) = n^2(n+1)$$
My work:
Base Case: $n = 1$
l.s = 2
r.s = 2
True
Induction Hypothesis: Assume for some $k$ that's an element of a natural number, $$1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ k(3k-1) = k^2(k+1)$$
Now show that, $$1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ (k+1)(3(k+1)-1) = (k+1)^2((k+1)+1)$$
$$(k+1)(3k +2) = (k+1)^2(k+2)
$$
$$3k^2 + 5k+ 2 = k^3 + 4k^2 + 5k + 2
$$
$$0 = k^3 + 4k^2-3k^2 + 5k - 5k + 2 - 2
$$
$$0 = k^3 + k^2
$$
by Induction Hypothesis, $k^3 + k^2 = k(3k-1)$
I know you're not supposed to start off with what you are trying to prove/show but if I reverse this whole process, wouldn't that be a correct proof? Is there a faster way? 
Thanks

Your proof is incredibly confusing (not least due to lack of proper type-setting).
As you know, you really should just start on one side and try to get to the other side:
$$\sum_{n=1}^{k+1}(n(3n-1)) = $$
$$\sum_{n=1}^{k}(n(3n-1)) + (k+1)(3(k+1)-1)= $$ (Inductive Hypothesis)
$$k^2(k+1) + (k+1)(3(k+1)-1)= $$
$$(k+1)(k^2 + 3(k+1) -1) =$$
$$(k+1)(k^2 +3k+2)=$$
$$(k+1)(k+1)(k+2)=$$
$$(k+2)(k + 1)^2$$
So it is really only in the second to last step that I was looking at the goal ($(k+2)(k + 1)^2$) that told me I had to factor out a $k+2$, but other than that I just worked from left to right
