Question

$\tt find \: the \: sum \: \frac{1}{1 + \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3} } + \frac{1}{ \sqrt{3} + \sqrt{4} } + ..upto 99terms.$
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Answer 1

We have to

$\tt find \: the \: sum \: \frac{1}{1 + \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3} } + \frac{1}{ \sqrt{3} + \sqrt{4} } + ..upto 99terms.$

Writing the rth term

The rth term of the given expression could be written as $\tt \frac{1}{\sqrt{r} + \sqrt{r+1}}$

Rationalising the den. of rth term :-

$\tt \frac{1}{\sqrt{r} + \sqrt{r+1}} = \frac{\sqrt{r+1} - \sqrt{r}}{(\sqrt{r} + \sqrt{r+1})(\sqrt{r+1} - \sqrt{r})} \\ \\ \\ = \frac{\sqrt{r+1} - \sqrt{r}}{{(\sqrt{r+1})}^2 - {(\sqrt{r})}^2} \\ \\ \\ = \frac{\sqrt{r+1} - \sqrt{r}}{(r+1) - (r)} \\ \\ \\ = \frac{\sqrt{r+1} - \sqrt{r}}{1} \\ \\ \\ = \sqrt{r+1} - \sqrt{r}$

Therefore, 1st term = $\sqrt{2} - \sqrt{1}$

2nd term = $\sqrt{3} - \sqrt{2}$

3rd term = $\sqrt{4} - \sqrt{3}$

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99th term = $\sqrt{100} - \sqrt{99}$

Adding all these term, we get

$\not{\sqrt{2}} - \sqrt{1} + \not{\sqrt{3}} - \not{\sqrt{2}} + \not{\sqrt{4}} - \not{\sqrt{3}} . . . . . . \not{\sqrt{99}} - \not{\sqrt{98}} + \sqrt{100} - \not{\sqrt{99}} \\ \\ \\ = \sqrt{100} - \sqrt{1} \\ \\ \\ = 10 - 1 \\ \\ \\ = 9$

Hence, the sum of these terms is 9.