Question

The value of $\sum _{i=1}^n\sum _{j=1}^i\sum _{k=1}^{j}1=220$ value if $n$ is.

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

Answer 1

The correct option is C $10$

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

${\sum }_{i=1}^n{\sum }_{j=1}^i(1+1+....jtimes)=220$

${\sum }_{i=1}^n{\sum }_{j=1}^i\left(j\right)=220$

${\sum }_{i=1}^n(1+2+3+.......+i)=220$

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

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

$\left(\frac{1}{2}\right)\left[\right(\frac{n(n+1)(2n+1)}{6})+(\frac{n(n+1)}{2}\left)\right]=220$

$\left(\frac{1}{2}\right)\left(\frac{n(n+1)\left[\right(2n+1)+3])}{6}\right)=220$

$n(n+1)(2n+4)=12\cdot 220$

$n(n+1)2(n+2)=12\cdot 220$

$n(n+1)(n+2)=6\cdot 220$

$n(n+1)(n+2)=6\cdot 2\cdot 10\cdot 11$

$n(n+1)(n+2)=10\cdot 11\cdot 12$

$\therefore n=10$