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Let $f (n) = [\frac{1}{5} + \frac{3 n}{100}] n,$ where $(x)$ denotes the greatest integer less than or equal to $x .$ Then $\sum_{n = 1}^{61} f (n)$

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Correct option is D

If 3n100<0.8 or n≤26,15+3n100=0If 0.8≤3n100<1.8, or 27≤n≤59then 15+3n100=1For n=60, 6115+3n100=2Thus, ∑n=161 f(n)=∑n=2759 n+∑n=6061 2n=12(33)(27+59)+2(60+61)=1661

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Let f(n)=15+3n100n, where (x) denotes the greatest integer less than or equal to x. Then ∑n=161 f(n)

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Let $f (n) = [\frac{1}{5} + \frac{3 n}{100}] n,$ where $(x)$ denotes the greatest integer less than or equal to $x .$ Then $\sum_{n = 1}^{61} f (n)$