Question

please help please. ....​

Answer 1

Answer :-

Explanation :-

$\\ \displaystyle \mathsf{S = \sqrt{4} + \frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2}} + ... + \frac{1}{\sqrt{196} + \sqrt{195}}} \\\\\\ \mathsf{S = \sqrt{4} + \frac{1}{\sqrt{2} + \sqrt{1}} + \frac{1}{\sqrt{3} + \sqrt{2}} + ... + \frac{1}{\sqrt{196} + \sqrt{195}}} \\\\\\ \mathsf{S = \sqrt{4} + \sum_{n = 2}^{196} \frac{1}{\sqrt{n} + \sqrt{n - 1}}}$

Rationalise,

$\displaystyle \mathtt{\frac{1}{\sqrt{n} + \sqrt{n - 1}} \cdot \frac{\sqrt{n} - \sqrt{n - 1}}{\sqrt{n} - \sqrt{n - 1}} = \frac{\sqrt{n} - \sqrt{n - 1}}{n - (n - 1)} = \boxed{\mathtt{\sqrt{n} - \sqrt{n - 1}}}}$

$\\ \displaystyle \mathsf{S = \sqrt{4} + \sum_{n = 2}^{196} \frac{1}{\sqrt{n} + \sqrt{n - 1}}} \\\\\\ \mathsf{S = 2 + \sum_{n = 2}^{196} \sqrt{n} - \sqrt{n - 1}} \\\\\\ \mathsf{S = 2 + \sum_{n = 2}^{196} \sqrt{n} - \sum_{n = 2}^{196} \sqrt{n - 1}} \\\\\\ \mathsf{S = 2 + \left ( \sqrt{2} + \sqrt{3} + ... + \sqrt{195} + \sqrt{196} \right ) - \left ( \sqrt{1} + \sqrt{2} + ... + \sqrt{195} \right )}$

$\\ \displaystyle \mathsf{S = 2 + \left ( \sqrt{2} + \sqrt{3} + ... + \sqrt{195} + \sqrt{196} \right ) - \left ( \sqrt{1} + \sqrt{2} + ... + \sqrt{195} \right )} \\\\ \mathsf{S = 2 + \left ( \sqrt{196} - \sqrt{1} \right )} \\\\ \mathsf{S = 2 + (14 - 1)} \\\\ \boxed{\boxed{\mathsf{S = 15}}}$