Question

If $y=\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{2^2}{1+x^4}+............+\frac{2^{100}}{1+x^{200}}$ , then find $y$ at $x=2$.

Answer 1

Explanation for the correct answer:

Step 1: Find the value of $y$.

Given,

$y=\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{2^2}{1+x^4}+............+\frac{2^{100}}{1+x^{200}}$

Now,

Add and subtract with $\frac{1}{1-x}$.

So,

$\begin{array}{rcl}y& =& \left(-\frac{1}{1-x}+\frac{1}{1-x}\right)+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{2^2}{1+x^4}+............+\frac{2^{100}}{1+x^{200}}\\ & =& -\frac{1}{1-x}+\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{2^2}{1+x^4}+............+\frac{2^{100}}{1+x^{200}}\\ & =& -\frac{1}{1-x}+\frac{2^2}{1-x^4}+\frac{2^2}{1+x^4}+............+\frac{2^{100}}{1+x^{200}}\\ & =& -\frac{1}{1-x}+\frac{2^{100}}{1-x^{200}}+\frac{2^{100}}{1+x^{200}}\\ & =& -\frac{1}{1-x}+\frac{2^{101}}{1-x^{400}}\end{array}$

Step 2: Find the value of $y$ at $x=2$:

Put $x=2$, in the value of $y$ obtained in Step 1.

$\begin{array}{rcl}y& =& -\frac{1}{1-\left(2\right)}+\frac{2^{101}}{1-{\left(2\right)}^{400}}\\ & =& 1+\frac{2^{101}}{1-2^{400}}\end{array}$

Hence, at $x=2$., $y=1+\frac{2^{101}}{1-2^{400}}$.