is equal to: [2026]

Question

$\sum_{n = 1}^{10} (\frac{528}{n (n + 1) (n + 2)})$ is equal to: [2026]

Smart Achievers · Accepted Answer

(2)

We have, $\sum_{n = 1}^{10} \frac{528}{n (n + 1) (n + 2)}$

Let, $T_{n} = \frac{1}{n (n + 1) (n + 2)}$

$= \frac{1}{2} [\frac{(n + 2) - n}{n (n + 1) (n + 2)}]$

$= \frac{1}{2} [\frac{1}{n (n + 1)} - \frac{1}{(n + 1) (n + 2)}]$

Now, $528 \sum_{n = 1}^{10} \frac{1}{2} [\frac{1}{n (n + 1)} - \frac{1}{(n + 1) (n + 2)}]$

$= \frac{528}{2} [\frac{1}{2} - \frac{1}{132}]$

$= \frac{528}{2} \times \frac{65}{132} = 130$

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