JEE MAIN - Mathematics (2023 - 13th April Evening Shift - No. 22)

Let $$[\alpha]$$ denote the greatest integer $$\leq \alpha$$. Then $$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots+[\sqrt{120}]$$ is equal to __________
Answer
825

Explanation

$$ \begin{aligned} & {[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots .+[120]} \\\\ & E=1+1+1+2+2+2+2+2+3+3+3+3+3 \\\\ & +3+3+4+4+\ldots \\\\ & E=3 \times 1+5 \times 2+7 \times 3+\ldots .+19 \times 9+10 \times 21 \\\\ & =\sum_{r=1}^{10}(2 r+1) r=2\left[\frac{10 \times 11 \times 21}{6}\right]+\frac{10 \times 11}{2} \\\\ & =770+55 \\\\ & =825 \\\\ & \end{aligned} $$

Comments (0)

Advertisement