Let f(n) = left[ frac{1}{3} + frac{3n}{100} r | vedclass

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(D) Given $f(n) = \left[ \frac{1}{3} + \frac{3n}{100} \right]n$. For $1 \le n \le 22$,$\frac{1}{3} + \frac{3n}{100} < \frac{1}{3} + \frac{66}{100} = 0

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$1399$

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(D) Given $f(n) = \left[ \frac{1}{3} + \frac{3n}{100} ight]n$. 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 $n = 56$,$\frac{1}{3} + \frac{3(56)}{100} = 0.333 + 1.68 = 2.013$. Thus,$\left[ \frac{1}{3} + \frac{3(56)}{100} ight] = 2$,so $f(56) = 2 	imes 56 = 112$. The sum is $\sum_{n=1}^{56} f(n) = \sum_{n=1}^{22} 0 + \sum_{n=23}^{55} n + f(56)$. $= 0 + \frac{(55-23+1)}{2}(23+55) + 112 = \frac{33}{2}(78) + 112 = 33 	imes 39 + 112 = 1287 + 112 = 1399$.

Let $f(n) = \left[ \frac{1}{3} + \frac{3n}{100} \right]n$,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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