How to prove $\sum\limits_{k=0}^n\ (3k^2+2k+1) = n^3 + 5 \begin{pmatrix}n+1\\2\end{pmatrix}+1$

Question

$\sum\limits_{k=0}^n\ (3k^2+2k+1) = n^3 + 5 \begin{pmatrix}n+1\\2\end{pmatrix}+1$

How would you go on proving this equation? Doesn't have to be induction..

Answer 1

Recall the following identities: $$\sum_{k=0}^n 1 = n+1$$ $$\sum_{k=0}^n k = \dfrac{n(n+1)}2$$ $$\sum_{k=0}^n k^2 = \dfrac{n(n+1)(2n+1)}6$$ And $$\sum_{k=0}^n (3k^2+2k+1) = 3 \sum_{k=0}^n k^2 + 2\sum_{k=0}^n k + \sum_{k=0}^n 1$$ Now plug in and obtain what you want.

Answer 2

$$(k+1)^3=k^3+3k^2+3k+1\quad=>\quad(k+1)^3-k^3=3k^2+3k+1\quad=>$$

$$=>\quad\sum_0^n(3k^2+2k+1)=\sum_0^n\Big[(3k^2+3k+1)-k\Big]=\sum_0^n\underbrace{(3k^2+2k+1)}_{\large(k+1)^3-k^3}-\sum_0^nk=$$

$$=(n+1)^3-\frac{n(n+1)}2=n^3+\underbrace{3n^2+3n}_{6\tfrac{n(n+1)}2}+1-\frac{n(n+1)}2=n^3+5\cdot\underbrace{\frac{n(n+1)}2}_{\Large{n+1\choose2}}+1.$$