Question

The value of $\sum _{i=1}^n\sum _{j=1}^i\sum _{k=1}^{j}1=220,$ then the value of $n$ equals

• A. $11$
• B. $12$
• C. $10$
• D. $9$

Answer 1

The correct option is C $10$

$\sum _{i=1}^n\sum _{j=1}^i\sum _{k=1}^{j}1=220$

$\Rightarrow \sum _{i=1}^n\sum _{j=1}^{i}j=220$

$\Rightarrow \sum _{i=1}^n\frac{i(i+1)}{2}=220$

$\Rightarrow \sum _{i=1}^n{i}^2+i=440$

$\Rightarrow \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=440$

$\Rightarrow \frac{n(n+1)}{2}[\frac{2n+1}{3}+1]=440$

$\Rightarrow \frac{n(n+1)(2n+4)}{2\times 3}=440$

$\Rightarrow n(n+1)(n+2)=1320$
$\Rightarrow 10\times 11\times 12$
$\Rightarrow n=10$