Sigma with pi 3

Algebra Level 1

$\sqrt{\left(\displaystyle\sum_{n=1}^{5} n^{2} - \displaystyle\sum_{h=2}^5 h^{2}\right) \displaystyle\prod_{k=3}^4 k^{2}}$ =

1 solution

Omkar Kulkarni
Jan 16, 2015

First we evaluate the bracket.

$\displaystyle \sum_{n=1}^{5} n^{2} - \displaystyle \sum_{h=2}^{5} h^{2}$

$= 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} - 2^{2} - 3^{2} - 4^{2} - 5^{2} = 1$

Therefore, the expression is simplified to

$\sqrt{\displaystyle \prod_{k=3}^{4} k^{2}}$

$= \displaystyle \prod_{k=3}^{4} k$

$= 3 \times 4 = \boxed {12}$