Follow the question the $f(n)=4n-1$ and $F(n)=\sum\_{k=0}^nf(k)$. And it ask you to write the recursive of $F(n)$. But I only know the recursive of $f(n)$ is 

$$f(n)=\begin{cases}
-1,&\text{if }n=0\\
-1+4(n-1)\text{ (maybe?)},&\text{if }n>0
\end{cases}$$

And find a closed-form for $F(n)$. And I know it's equal to $2n^2+n-1$, but I don't know how to explain how I find it...QWQ

Please help me thx you!!