Let f(n) = [frac{1}{3} + frac{3n}{100}]n,wher | vedclass

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(D) Given $f(n) = [\frac{1}{3} + \frac{3n}{100}]n$,where $[x]$ is the greatest integer function.For $1 \le n \le 22$,$\frac{1}{3} + \frac{3n}{100} < \

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$1399$

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(D) Given $f(n) = [\frac{1}{3} + \frac{3n}{100}]n$,where $[x]$ is the greatest integer function.For $1 \le n \le 22$,$\frac{1}{3} + \frac{3n}{100} For $23 \le n \le 55$,$1 \le \frac{1}{3} + \frac{3n}{100} For $n = 56$,$\frac{1}{3} + \frac{3(56)}{100} = 0.333 + 1.68 = 2.013$,so $f(56) = 2 \cdot 56 = 112$.Therefore,$\sum_{n=1}^{56} f(n) = \sum_{n=1}^{22} 0 + \sum_{n=23}^{55} n + 112$.The sum $\sum_{n=23}^{55} n$ is an arithmetic progression with $33$ terms: $\frac{33}{2}(23 + 55) = \frac{33}{2}(78) = 33 \cdot 39 = 1287$.Thus,the total sum is $1287 + 112 = 1399$.

Let $f(n) = [\frac{1}{3} + \frac{3n}{100}]n$,where $[x]$ denotes the greatest integer less than or equal to $x$. Then $\sum_{n=1}^{56} f(n)$ is equal to

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