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(B) We need to calculate the sum $S = \sum_{n=1}^{120} [\sqrt{n}]$.For a given integer $k$,$[\sqrt{n}] = k$ when $k^2 \leq n < (k+1)^2$.The number of

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$825$

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(B) We need to calculate the sum $S = \sum_{n=1}^{120} [\sqrt{n}]$.For a given integer $k$,$[\sqrt{n}] = k$ when $k^2 \leq n The number of such integers $n$ is $(k+1)^2 - k^2 = 2k+1$.We observe that for $k=1, 2, \ldots, 10$,the values of $n$ range from $1$ to $120$ because $10^2 = 100$ and $11^2 = 121$.For $k=1, 2, \ldots, 9$,the number of terms is $2k+1$.For $k=10$,the range is $100 \leq n \leq 120$,which gives $120 - 100 + 1 = 21$ terms.Sum $S = \sum_{k=1}^{9} k(2k+1) + 10(21)$.$S = \sum_{k=1}^{9} (2k^2 + k) + 210$.$S = 2 	imes \frac{9(10)(19)}{6} + \frac{9(10)}{2} + 210$.$S = 570 + 45 + 210 = 825$.

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

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