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Question

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

A

$\frac{n\left(n+1\right)}{2}$

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B

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

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C

$\frac{n\left(n+1{\right)}^{2}}{2}$

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D

None of these

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Solution

## The correct option is B $\frac{n\left(n+1\right)\left(n+2\right)}{6}$Explanation for the correct answer:Simplifying the given expression:$\underset{i=1}{\overset{n}{⇒\sum }}\sum _{j=1}^{1}\sum _{k=1}^{j}1\phantom{\rule{0ex}{0ex}}⇒\sum _{i=1}^{n}\sum _{j=1}^{1}j\left[\because \sum _{k=1}^{j}1=1\right]\phantom{\rule{0ex}{0ex}}⇒\sum _{i=1}^{n}\left(1+2+3+.....+i\right)\left[\text{Consider}j=1+2+3+...+i\right]\phantom{\rule{0ex}{0ex}}⇒\sum _{i=1}^{n}\left(\frac{i\left(i+1\right)}{2}\right)\left[\because \text{sumofnnaturalnumbersis}\frac{n\left(n+1\right)}{2}\right]\phantom{\rule{0ex}{0ex}}⇒\sum _{i=1}^{n}\left(\frac{{i}^{2}+i}{2}\right)\phantom{\rule{0ex}{0ex}}⇒\left[\frac{n\left(n+1\right)\left(2n+1\right)}{12}\right]+\left[\frac{n\left(n+1\right)}{4}\right]\left[\because \text{sumofsquaresofnnaturalnumbers}=\frac{n\left(n+1\right)\left(2n+1\right)}{2}\right]\phantom{\rule{0ex}{0ex}}⇒\left(\frac{n\left(n+1\right)}{4}\right)\left[\left(\frac{2n+1}{3}\right)+1\right]\phantom{\rule{0ex}{0ex}}⇒\frac{n\left(n+1\right)\left(2n+4\right)}{4×3}\phantom{\rule{0ex}{0ex}}⇒\frac{n\left(n+1\right)\left(n+2\right)}{6}$Thus, $\sum _{i=1}^{n}\sum _{j=1}^{1}\sum _{k=1}^{j}1=\frac{n\left(n+1\right)\left(n+2\right)}{6}$Hence, option (B) is correct.

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