Calculate $\sum_{j=1}^{n}\sum_{i=1}^{j}j\cdot i$ using binomial coefficients.

Question

I want to check my ideas to calculate $\sum_{j=1}^{n}\sum_{i=1}^{j}j\cdot i$.

I exploited the hockey stick identity in the following way:

$$\begin{eqnarray*}\sum_{j=1}^{n}\binom{j}{1}\cdot \sum_{i=1}^{j} \binom{i}{1}&=&\sum_{j=1}^{n}\binom{j}{1}\cdot \binom{j+1}{2}\\&=&3\left[\sum_{j=1}^{n}\left(\frac{j+2}{3}\cdot \binom{j+1}{2}-\frac{2}{3}\binom{j+1}{2}\right)\right]\\&=&3\binom{n+2}{4}+\binom{n+2}{3}.\end{eqnarray*}$$

Thanks everybody.

Answer 1

That is fine but there is a faster way through symmetry: $$ \sum_{j=1}^{n}\sum_{i=1}^{j}ij = \sum_{1\leq i\leq j\leq n}ij = \frac{1}{2}\left[\left(\sum_{i=1}^{n}i\right)^2+\sum_{i=1}^{n}i^2\right]=\frac{n^2(n+1)^2}{8}+\frac{n(n+1)(2n+1)}{12}.$$