Question

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

• A. $\frac{n(n+1)}{2}$
• B. $\frac{n(n+1)(n+2)}{6}$
• C. $\frac{n(n+1{)}^2}{2}$
• D. None of these

Answer 1

The correct option is B

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

Explanation for the correct answer:

Simplifying the given expression:

$$\begin{gathered} \underset{i=1}{\overset{n}{\Rightarrow \sum }}\sum _{j=1}^1\sum _{k=1}^{j}1 \\ \Rightarrow \sum _{i=1}^n\sum _{j=1}^{1}j\left[\because \sum _{k=1}^{j}1=1\right] \\ \Rightarrow \sum _{i=1}^n(1+2+3+.....+i)[\text{Consider}j=1+2+3+...+i] \\ \Rightarrow \sum _{i=1}^n\left(\frac{i(i+1)}{2}\right)\left[\because \text{sum of n natural numbers is}\frac{n(n+1)}{2}\right] \\ \Rightarrow \sum _{i=1}^n\left(\frac{i^2+i}{2}\right) \\ \Rightarrow \left[\frac{n(n+1)(2n+1)}{12}\right]+\left[\frac{n(n+1)}{4}\right]\left[\because \text{sum of squares of n natural numbers}=\frac{n(n+1)(2n+1)}{2}\right] \\ \Rightarrow \left(\frac{n(n+1)}{4}\right)\left[\left(\frac{2n+1}{3}\right)+1\right] \\ \Rightarrow \frac{n(n+1)(2n+4)}{4\times 3} \\ \Rightarrow \frac{n(n+1)(n+2)}{6} \end{gathered}$$

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

Hence, option (B) is correct.