Let me type the whole question:
Prove that $1\cdot2+2\cdot5+3\cdot8+ \cdots+n(3n-1)= n^2(n+1)$ for $n\in\mathbb N$
If you have a link for online-documentary for induction methods and examples please post it here it would be really useful.
Thanks in advance.
On
Let P(n) be the conjecture, $$\sum_{k+1}^{n}(k(3k-1))=n^{2}(n+1)$$
Assume true for n = 1 $$ 1 * 2 = 2 $$ $$ 1^{2}*(1+1) =2 $$ Therefore, P(1) is true
Assume true for P(i), for n = i
$$\sum_{k=1}^{i}k(3k-1) = i^{2}*(i+1) $$
Show that conjecture is true for P(i+1) for n = i+1 P(i+1) = P(i) + U(i+1), where U(n) is the nth term
$$i^{2}*(i+1) + (i+1)(3(i+1)-1)$$ $$=(i+1)*(i^{2}+3i+2) $$ $$=(i+1)(i+2)(i+1)$$ $$=(i+1)^{2}(i+2) $$ $$=(i+1)^{2}((i+1)+1)$$
Hence, P(1) and P(i+1) are true when P(i) is assumed to be true. Hence, the conjecture is true.
It is sometimes easier to start by the end
$$(n+1)^2(n+2)-n^2(n+1)=$$
$$(n+1)\left((n+1)(n+2)-n^2\right)=$$
$$(n+1)(n^2+n+2n+2-n^2)=$$
$$(n+1)\left(3(n+1)-1\right)$$
qed.