Question

The value of $\sum _{i=1}^n\sum _{j=1}^i\sum _{k=1}^{j}1=$

Answer 1

${\sum }_{i=1}^n{\sum }_{j=1}^i{\sum }_{k=1}^{j}1={\sum }_{i=1}^n{\sum }_{j=1}^{i}j={\sum }_{i=1}^n(1+2+3+......+11i)={\sum }_{i=1}^n\left(\frac{i(i+1)}{2}\right)=\sum \left(\frac{\left(i^2\right)}{2}\right)+\sum i=\left(\frac{n(n+1)(2n+1)}{12}\right)+\left(\frac{n(n+1)}{2}\right)=\left(\frac{n(n+1)}{2}\right)\left[\right(\frac{(2n+1)+6}{6}\left)\right]=\left(\frac{n(n+1)(2n+7)}{12}\right)$