First slide

Special Series in Sequences and Series

Question

$The value of \sqrt{1 + \frac{1}{1^{2}} + \frac{1}{2^{2}}} + \sqrt{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + . . + \sqrt{1 + \frac{1}{(2020)^{2}} + \frac{1}{(2021)^{2}}} is$

Difficult

A

$2021$
B

$2021 - \frac{1}{2021}$
C

$2020 - \frac{1}{2020}$
D

$2021 - \frac{1}{2020}$

Solution

$\begin{array}{l} T_{n} = \sqrt{1 + \frac{1}{n^{2}} + \frac{1}{(n + 1)^{2}}} = \sqrt{\frac{n^{2} + (n + 1)^{2} + n^{2} (n + 1)^{2}}{n^{2} (n + 1)^{2}}} \\ = \sqrt{\frac{2 n^{2} + 2 n + 1 + n^{2} (n + 1)^{2}}{n^{2} (n + 1)^{2}}} \\ = \frac{n (n + 1) + 1}{n (n + 1)} = 1 + \frac{1}{n} - \frac{1}{n + 1} \\ r e q u i r e d s u m = S = \sum_{x = 1}^{2020} (1 + \frac{1}{n} - \frac{1}{n + 1}) = 2020 + 1 - \frac{1}{2021} = 2021 - \frac{1}{2021} \end{array}$

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