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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