Question

Find the sum $\sum _{i=1}^n\sum _{j=1}^i\sum _{k=1}^{j}1$.

• A. $\frac{n(n+1)(4n+5)}{24}$
• B. $\frac{n(n+1)(2n+1)}{6}$
• C. $\frac{n(n+1)(2n+1)}{24}$
• D. None of These

Answer 1

The correct option is A $\frac{n(n+1)(4n+5)}{24}$

$S=\sum _{i=1}^n\sum _{j=1}^i\sum _{k=1}^{j}1=\sum _{i=1}^n\sum _{j=1}^{i}j=\sum _{i=1}^n\frac{i(i+1)}{2}=\frac{1}{2}\sum _{i=1}^n(i^2+i)=\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}]=\frac{n(n+1)(4n+5)}{24}$
Ans: A