Question

Sum the series $\frac{1}{1+\sqrt{x}},\frac{1}{1-x},\frac{1}{1-\sqrt{x}}$, .... to $n$ terms.

Answer 1

Terms of the series are:

$\frac{1}{1+\sqrt{x}},\frac{1}{1-x},\frac{1}{1-\sqrt{x}}$

They can also be written as:

$\frac{1-\sqrt{x}}{1-x},\frac{1}{1-x},\frac{1+\sqrt{x}}{1-x}$

$k^{th}$ term of the series $=\frac{1+(k-2)\sqrt{x}}{1-x}$

Sum of $n$ terms $=\sum _{k=1}^n\frac{1+(k-2)\sqrt{x}}{1-x}$

$=\frac{1}{1-x}\times (\sum _{k=1}^{n}1+\sqrt{x}\sum _{k=1}^n(k-2))$

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

$=\frac{1}{1-x}\times (n+\sqrt{x}\left(\frac{n^2+n-4n}{2}\right))$

$=\frac{1}{2(1-x)}\times (2n+\sqrt{x}(n^2-3n\left)\right)$

$=\frac{n}{2(1-x)}(2+(n-3\left)\sqrt{x}\right)$