Question

If {x} is the least integer greater than or equal to x, then find the value of the following series:
$\left\{\sqrt{1}\right\}+\left\{\sqrt{2}\right\}+\left\{\sqrt{3}\right\}+\left\{\sqrt{4}\right\}+...+\left\{\sqrt{99}\right\}+\left\{\sqrt{100}\right\}$

• A. 715
• B. 55
• C. 157
• D. 835

Answer 1

The correct option is A 715

Let $S=\left\{\sqrt{1}\right\}+\left\{\sqrt{2}\right\}+\left\{\sqrt{3}\right\}+\left\{\sqrt{4}\right\}+...+\left\{\sqrt{99}\right\}+\left\{\sqrt{100}\right\}\Rightarrow S=1+2+2+2+3+3+3+3+3+...+10\Rightarrow S=1+3\left(2\right)+5\left(3\right)+7\left(4\right)+9\left(5\right)+...+19\left(10\right)$
$Since{T}_n=(2n-1)n=2n^2-n\therefore S_n=\sum (2n^2-n)=2\sum n^2-\sum n$
$=\frac{2n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}=\frac{n(n+1)}{2}\left[\frac{2}{3}\right(2n+1)-1]=\frac{n(n+1)}{2}\left[\frac{4n+2-3}{3}\right]=\frac{n(n+1)(4n-1)}{6}\therefore S_{10}=\frac{10\times 11\times 39}{6}=715$