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) $\frac{1}{2}|\sqrt{2 n+1}-1|$
(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:
$S_{n}=\sum_{k=1}^{n} T_{k}$
$=\sum_{k=1}^{n}\left(\frac{\sqrt{2 k+1}-\sqrt{2 k-1}}{2}\right)$
$=\frac{1}{2} \sum_{k=1}^{n}(\sqrt{2 k+1}-\sqrt{2 k-1})$
$=\frac{1}{2}[(\sqrt{3}-\sqrt{1})+(\sqrt{5}-\sqrt{3})+(\sqrt{7}-\sqrt{5})+\ldots+(\sqrt{2 n+1}-\sqrt{2 n-1})]$
$=\frac{1}{2}\{(-1)+\sqrt{2 n+1}\}$
$=\frac{1}{2}\{\sqrt{2 n+1}-1\}$