Prove using mathematical induction pt 2 Assumed that i asked a question like 30 min ago thinking i got the hang of this, seems not.
So $$1^2+4^2+7^2+\dots+(3n-2)^2=\frac12n(6n^2-3n-1) \text{ for all } n\in\mathbb N$$
This time it seems way harder with the squares.
so i did the steps and got stuck on the 3rd step(Again).
Step 1: prove LHS = RHS which it does for n=1
Step 2: Assume $n=k$ is true $$1^2+4^2+7^2+\dots+(3k-2)^2=\frac12k(6k^2-3k-1)$$
Step 3: would $n = k+1$? And would $n = k+1$ work for all equations?could someone help me with the last step, would be appreciated thanks
EDIT: Cheers for the help, i know where i went wrong!
 A: Well... $(3 - 2)^2 = 1 = \frac{6-3-1}{2}$ so it is true for $n=1$. Now, suppose it's true for $n=k$. Then, $1^2+4^2+\dots+(3k-2)^2+(3k+1)^2 = \frac{k\cdot(6k^2-3k-1)}{2} + (3k+1)^2 =\cdots$ 
A: Hypothesis:
$$
1^2+4^2+7^2+\dots+(3n-2)^2=\frac{1}{2}n(6n^2-3n-1)
$$
Thesis:
$$
1^2+4^2+7^2+\dots+(3n-2)^2+(3(n+1)-2)^2=\frac{1}{2}(n+1)(6(n+1)^2-3(n+1)-1)
$$
By the induction hypothesis
$$
1^2+4^2+7^2+\dots+(3n-2)^2+(3(n+1)-2)^2=
\frac{1}{2}n(6n^2-3n-1)+(3(n+1)-2)^2
$$
Write this as a polynomial in $n$; write
$$
\frac{1}{2}(n+1)(6(n+1)^2-3(n+1)-1)
$$
as a polynomial in $n$.
Verify that the expressions are the same. End.
A: $$1^2+4^2+7^2+\dots+(3n-2)^2=\frac{1}{2}n(6n^2-3n-1), \forall n\in N$$
$$1^2+4^2+7^2+\dots+(3n-2)^2+(3n+1)^2=\frac{1}{2}n(6n^2-3n-1))+(3n+1)^2 $$
$$=\frac{1}{2}(6n^3-3n^2-n+2(3n+1)^2)=\frac{1}{2}(6n^3+15n^2+11n+2))= $$
$$\frac{1}{2}(6n^3+12n^2+6n+3n^2+5n+2)=$$
$$\frac{1}{2}(6n(n^2+2n+1)+3n^2+3n+2n+2)=$$
$$\frac{1}{2}(6n(n+1)^2+3n(n+1)+2(n+1))=$$
$$\frac{1}{2}(n+1)(6n(n+1)+3n+2)=\frac{1}{2}(n+1)(6n^2+9n+2)=$$
$$=\frac{1}{2}(n+1)(6n^2+12n+6-3n-3-1)=$$
$$=\frac{1}{2}(n+1)(6(n+1)^2-3(n+1)-1)$$
