The sum to n terms of the series
Question:

The sum to $n$ terms of the series $\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\ldots+\ldots$ is

(a) $\sqrt{2 n+1}$

(b) $\frac{1}{2} \sqrt{2 n+1}$

(c) $\sqrt{2 n+1}-1$

(d) $\frac{1}{2}|\sqrt{2 n+1}-1|$

Solution:

(d) $\frac{1}{2}|\sqrt{2 n+1}-1|$

Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$T_{n}=\frac{1}{\sqrt{2 n-1}+\sqrt{2 n+1}}=\frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{2}$

Now,

Let $S_{n}$ be the sum of $n$ terms of the given series.

Thus, we have:

$S_{n}=\sum_{k=1}^{n} T_{k}$

$=\sum_{k=1}^{n}\left(\frac{\sqrt{2 k+1}-\sqrt{2 k-1}}{2}\right)$

$=\frac{1}{2} \sum_{k=1}^{n}(\sqrt{2 k+1}-\sqrt{2 k-1})$

$=\frac{1}{2}[(\sqrt{3}-\sqrt{1})+(\sqrt{5}-\sqrt{3})+(\sqrt{7}-\sqrt{5})+\ldots+(\sqrt{2 n+1}-\sqrt{2 n-1})]$

$=\frac{1}{2}\{(-1)+\sqrt{2 n+1}\}$

$=\frac{1}{2}\{\sqrt{2 n+1}-1\}$

 

 

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