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

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

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

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

Answer 1

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

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:

$$\begin{gathered} S_n=\sum _{k=1}^n{T}_k \\ =\sum _{k=1}^n\left(\frac{\sqrt{2k+1}-\sqrt{2k-1}}{2}\right) \\ =\frac{1}{2}\sum _{k=1}^n\left(\sqrt{2k+1}-\sqrt{2k-1}\right) \\ =\frac{1}{2}\left[\left(\sqrt{3}-\sqrt{1}\right)+\left(\sqrt{5}-\sqrt{3}\right)+\left(\sqrt{7}-\sqrt{5}\right)+...+\left(\sqrt{2n+1}-\sqrt{2n-1}\right)\right] \\ =\frac{1}{2}\left\{\left(-1\right)+\sqrt{2n+1}\right\} \\ =\frac{1}{2}\left\{\sqrt{2n+1}-1\right\} \end{gathered}$$