Prove by induction that $n^2 + n + 1 \forall n\geq 1$ given the following recurrence relation My question is as follows:
Consider the following recurrence relation:
$a_{n} = a_{n-1}+2n$, with $a_{1}=3$
Prove by induction that
$a_{n}=n^{2}+n+1 \forall  n\geq 1$
I have no idea how to even approach this question. I am quite familiar with the process of induction but am just unsure how to draw that in for this question? I am not very familiar with recurrence relations.

Induction is usually achieved in three steps.
1) Initialization. You just need to check that your property is valid for the lowest integer required (here $n = 1$). So what you need to check is that indeed, $a_1 = 1^2 + 1 + 1$.
2) Heredity. This step is usually the most difficult, and you have to be careful when writing it. At this point, you assume that your property is valid for some $n \geq 1$ (and certainly not for all $n$, in which case there is nothing left to show).
Therefore, you assume that, for some $n \geq 1$, $a_n = n^2 + n + 1$. Now use the recurrence relation to deduce that $a_{n+1} = (n+1)^2 + (n+1) + 1$. 
3) Conclusion. In step 2, you have shown that if your property is valid at rank $n$, it is also valid at rank $n+1$. In step 1, you have shown that the property is valid at rank 1. Thus, the property is valid at rank 1+1 = 2, then at rank 2+1 = 3,... and finally for any $n \geq 1$. This step is usually skipped once one is familiar with the whole induction process. 
