Question

If [X] is the greatest integer less than or equal to x, then find the value of the following series $\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]+\left[\sqrt{4}\right]+...+\left[\sqrt{323}\right]$

• A. 3237
• B. 2373
• C. 3723
• D. None of these

Answer 1

The correct option is C 3723

Let $S=\left[\sqrt{1}\right]+[\sqrt{2}+[\sqrt{3}]+...+[\sqrt{323}]\Rightarrow S=1+1+1+2+2+2+2+2+...+17+17\Rightarrow S=1(3)+2(5)+3(7)+4(9)+...+17(35)\because T_n=n(2n+1)\therefore S_n=\sum n(2n+1)=2\sum n^2+\sum n=\frac{2.n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)}{2}\left[\frac{2}{3}\right(2n+1)+1]=\frac{n(n+1)}{2}\left[\frac{4n+5}{3}\right]=\frac{n(n+1)(4n+5)}{6}\therefore S=\frac{17\times 18\times 73}{6}=3723$.