Sum of the series P=121+2+132+23+…+110099+99100 is

A
$1 / 10$
B
$3 / 10$
C
$9 / 10$
D
$1 / 2$

Solution:

Let $t_{r}$ denote the $r t h$ term of the series, then
$\begin{array}{l} t_{r} = \frac{1}{(r + 1) \sqrt{r} + r \sqrt{r + 1}} = \frac{1}{\sqrt{r (r + 1)}} \frac{1}{\sqrt{r + 1} + \sqrt{r}} \\ = \frac{\sqrt{r + 1} - \sqrt{r}}{\sqrt{r (r + 1)}} = \frac{1}{\sqrt{r}} - \frac{1}{\sqrt{r + 1}} \end{array}$
Thus, $P = \sum_{r = 1}^{99} [\frac{1}{\sqrt{r}} - \frac{1}{\sqrt{r + 1}}] = 1 - \frac{1}{\sqrt{100}} = \frac{9}{10}$

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Sum of the series $P = \frac{1}{2 \sqrt{1} + \sqrt{2}} + \frac{1}{3 \sqrt{2} + 2 \sqrt{3}} + \dots + \frac{1}{100 \sqrt{99} + 99 \sqrt{100}}$ is

Solution:

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