Question

The sum to n terms of the series \[\frac{1}{\sqrt{1} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + . . . . + . . . .\]  is

पर्याय

• \[\sqrt{2n + 1}\]

• \[\frac{1}{2}\sqrt{2n + 1}\]

• \[\sqrt{2n + 1} - 1\]

• \[\frac{1}{2}\left\{ \sqrt{2n + 1} - 1 \right\}\]

   

Answer 1

\[\frac{1}{2}\left\{ \sqrt{2n + 1} - 1 \right\}\]

Let \[T_n\] be the nth term of the given series.
Thus, we have:

\[T_n = \frac{1}{\sqrt{2n - 1} + \sqrt{2n + 1}} = \frac{\sqrt{2n + 1} - \sqrt{2n - 1}}{2}\]

Now,
Let \[S_n\]  be the sum of n terms of the given series.

Thus, we have:

\[S_n = \sum^n_{k = 1} T_k \]

\[ = \sum^n_{k = 1} \left( \frac{\sqrt{2k + 1} - \sqrt{2k - 1}}{2} \right)\]

\[ = \frac{1}{2} \sum^n_{k = 1} \left( \sqrt{2k + 1} - \sqrt{2k - 1} \right)\]

\[ = \frac{1}{2}\left[ \left( \sqrt{3} - \sqrt{1} \right) + \left( \sqrt{5} - \sqrt{3} \right) + \left( \sqrt{7} - \sqrt{5} \right) + . . . + \left( \sqrt{2n + 1} - \sqrt{2n - 1} \right) \right]\]

\[ = \frac{1}{2}\left\{ \left( - 1 \right) + \sqrt{2n + 1} \right\}\]

\[ = \frac{1}{2}\left\{ \sqrt{2n + 1} - 1 \right\}\]