Question

$\sum _{i=1}^n\sum _{j=1}^i\sum _{k=1}^{j}1$ is equal to

• A. $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$
• B. ${\left(\frac{n\left(n+1\right)}{2}\right)}^2$
• C. $\frac{n\left(n+1\right)}{2}$
• D. $\frac{n\left(n+1\right)\left(n+2\right)}{6}$

Answer 1

The correct option is A $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

${\sum }_{i=1}^n{\sum }_{j=1}^i{\sum }_{k=1}^{j}1=\frac{n(n+1)(2n+1)}{6}$