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