Question about a sum Why is it that
$$\sum_{k=1}^n(n+k+1)(n+1)=\frac{3}{2}\sum_{k=1}^n3k^2+k.$$
I cannot understand it. This is not homework, I am just a little interested in this!
 A: These sums are classic and we can prove it by induction
$$\sum_{k=1}^n k=\frac{n(n+1)}{2}$$
and
$$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$$
so
$$\frac{3}{2}\sum_{k=1}^n3k^2+k=\frac{3n(n+1)^2}{2}$$
and on the other way
$$\sum_{k=1}^n(n+k+1)(n+1)=(n+1)\sum_{k=1}^n(n+k+1)\\=(n+1)\left(n(n+1)+\sum_{k=1}^nk\right)=\frac{3n(n+1)^2}{2}$$
so the equality is proved.
A: Just expand. $(n+k+1)(n+1) = n^2+n+kn+k+n+1 = n^2+n(k+2)+k+1$ so
$\sum_{k = 1}^n n^2+n(k+2)+k+1 = \sum_k n^2+\sum_k n(k+2)+\sum_k k + \sum_k 1$.
In the first term, $n$ is constant so the sum is just $n^2 \cdot n$, etc. Do this for all the sums and compare the two sides. 
A: The first sum can be written as$$\sum_{k=1}^n(n+k+1)(n+1)=(n+1)\sum_{k=1}^n(n+k+1)=(n+1)n(n+1)+(n+1)\sum_{k=1}^nk\\=n(n+1)^2+\frac{n(n+1)^2}{2}=\frac{3n(n+1)^2}{2}$$
instead the second one as
$$\frac{3}{2}\sum_{k=1}^n3k^2+k=\frac{3}{2}\left(\sum_{k=1}^n3k^2+\sum_{k=1}^nk\right)=\frac{3}{2}\left(3\sum_{k=1}^nk^2+\sum_{k=1}^nk\right)=\frac{3}{2}\left(3\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right)=\frac{3}{2}\left(\frac{n(n+1)}{2}(2n+2)\right)=\frac{3}{2}\left(\frac{n(n+1)}{2}2(n+1)\right)=\frac{3n(n+1)^2}{2}$$
which equals the first one!
