Question

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

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

Answer 1

Answer: (D)

$\frac{n(n+1)(n+2)}{6}$

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