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

The claim is as follows:
$$\sum_{k=1}^{n}k(3k-1)=n^2(n+1)$$
So consider the base case, that is, when $n=1$
Then:
$$\sum_{k=1}^{1}k(3k-1)=1(3(1)-1)=2 \ \checkmark$$
Assume the claim holds for the $n^{th}$ case, that is:
$$\sum_{k=1}^{n}k(3k-1)=n^2(n+1)$$
Then show it holds for the $(n+1)^{th}$ case:
$$\sum_{k=1}^{n+1}k(3k-1)=(n+1)^2(n+2)$$
So consider the following:
$$\sum_{k=1}^{n+1}k(3k-1)=\sum_{k=1}^{n}k(3k-1)+(n+1)(3(n+1)-1)=\sum_{k=1}^{n}k(3k-1)+(n+1)(3n+2)$$
Substituting our assumption, we have:
$$=n^2(n+1)+(n+1)(3n+2)=(n+1)(n^2+3n+2)=(n+1)(n+1)(n+2)=(n+1)^2(n+2)$$
Thus the $(n+1)^{th}$ case holds, given the $n^{th}$ case.
Therefore by induction, the original claim holds.
