Question

Let [x] denote the greatest integer part of a real number x. If
$M={\sum }_{n=1}^{40}\left[\frac{n^2}{2}\right]$
then m equals

• A. $5700$
• B. $5720$
• C. $5740$
• D. $11060$

Answer 1

The correct option is D $11060$

$M=\sum _{r=1}^{40}\left[\frac{n^2}{2}\right]$
Splitng n in odd number and in even number

$M=\sum _{x=1}^{20}\left[\frac{(2x{)}^2}{2}\right]+\sum _{y=0}^{19}\left[\frac{(2y+1{)}^2}{2}\right]$

$M=\sum _{x=1}^{20}\left[2x^2\right]+\sum _{y=0}^{19}[2y^2+2y+\frac{1}{2}]$

$=\sum _{x=1}^{20}2x^2+\sum _{y=0}^{19}2y^2+2y$
Now $\sum _{x=1}^{20}2x^2=\frac{2\times 20(20+1)\left(41\right)}{6}$ and $\sum _{y=0}^{19}(2y^2+2y)=\frac{2\times 19\times (19+1)\left(39\right)}{6}+\frac{2\times 19\times 20}{2}$

$2x\left(\frac{20\times 21\times 41}{6}\right)+2x\frac{(19\times 20\times 39)}{6}+2\frac{(19\times 20)}{2}$

$5740+4940+380=11060$