Question about a sum

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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!