Medium INTEGER +4 / -1 PYQ · JEE Mains 2023

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

Answer (integer) 825

Solution

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

About this question

Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane

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