First slide

Special Series in Sequences and Series

Question

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

Moderate

A

2013
B

1869
C

1947
D

1661

Solution

If $\frac{3 n}{100} < 0.8$ or $n \leq 26,$
$[\frac{1}{5} + \frac{3 n}{100}] = 0$
If $0.8 \leq \frac{3 n}{100} < 1.8,$ or $27 \leq n \leq 59$
then $[\frac{1}{5} + \frac{3 n}{100}] = 1$
For $n = 60, 61 [\frac{1}{5} + \frac{3 n}{100}] = 2$
Thus, $\sum_{n = 1}^{61} f (n) = \sum_{n = 27}^{59} n + \sum_{n = 60}^{61} 2 n$
$\begin{array}{l} = \frac{1}{2} (33) (27 + 59) + 2 (60 + 61) \\ = 1661 \end{array}$

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