First slide

Sigma operation series

Question

If $\sum_{k = 1}^{n} ϕ (k) = \frac{2 n}{n + 1},$ then $\sum_{k = 1}^{10} \frac{1}{ϕ (k)}$ is equal to

Moderate

A

$\frac{11}{20}$
B

220
C

$\frac{55}{18}$
D

110

Solution

Let $S_{n} = \sum_{k = 1}^{n} ϕ (k) = \frac{2 n}{n + 1}$
$\begin{array}{l} \Rightarrow ϕ (n) = S_{n} - S_{n - 1} = \frac{2 n}{n + 1} - \frac{2 (n - 1)}{n} \\ = \frac{2 [n^{2} - (n^{2} - 1)]}{(n + 1) n} = \frac{2}{n (n + 1)} \\ \Rightarrow \frac{1}{ϕ (n)} = \frac{n (n + 1)}{2} \\ \Rightarrow \sum_{k = 1}^{10} \frac{1}{ϕ (n)} = \frac{1}{2} \sum_{k = 1}^{10} (k^{2} + k) \\ = \frac{1}{2} \cdot \frac{1}{6} (10) (11) (21) + \frac{1}{2} \cdot \frac{1}{2} (10) (11) \\ = 220 \end{array}$

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