Solution to recurrence relation, as a formula involving summation operator

Question

Here is what I am tasked with..

Find a solution to the recurrence relation:

$F(0) = 2$

$F(n+1) = F(n) + 2n^2 - 1$

as a formula involving the summation operator

$$\sum_{i=1}^n$$

Sorry for the wonky formatting. Anyway, I am mostly looking for pointers/hints as to where or how to begin as I know it is heavily encouraged to try to work it out and not just ask for the answer. I have worked a little with finding closed-form solutions to factorial, fibonacci etc. But this problem states to write as a summation, not a closed form solution. Thanks!

Answer 1

Hint:(Use telescoping)

$F(n+1) -F(n) =(2n^2 -1)$

...

$F(1) -F(0)=(2\cdot 0^2 -1)$

Now add these up.

Can you take it from here?

Answer 2

Note that $F(n+1)-F(n)=2n^2-1$. Then $\sum_{n=0}^k 2n^2-1=\sum_{n=0}^kF(n+1)-F(n)$. But in the RHS:

$$\sum_{n=0}^kF(n+1)-F(n)=\sum_{n=0}^kF(n+1)-\sum_{n=0}^kF(n)\\ =F(k+1)+\sum_{n=0}^{k-1}F(n+1)-F(0)-\sum_{n=1}^kF(n)\\ =F(k+1)-F(0)+\sum_{n=0}^{k-1}F(n+1)-\sum_{n=0}^{k-1}F(n+1)\\ =F(k+1)-F(0)\\ =F(k+1)-2$$.

So, $F(k+1)=\sum_{n=0}^k(2n^2-1)+2$