Question

Simplify the equation given below.

Answer 1

Answer:

Step-by-step explanation:

We have,

$\dfrac{1}{a+1}+\dfrac{2}{a^2+1}+\dfrac{4}{a^4+1}+\dfrac{8}{a^8-1}$

$=\dfrac{1}{a+1}+\dfrac{2}{a^2+1}+\dfrac{4}{a^4+1}+\dfrac{8}{(a^4-1)(a^4+1)}$

$=\dfrac{1}{a+1}+\dfrac{2}{a^2+1}+\dfrac{4}{a^4+1}+4\cdot\dfrac{1+1}{(a^4-1)(a^4+1)}$

$=\dfrac{1}{a+1}+\dfrac{2}{a^2+1}+\dfrac{4}{a^4+1}+4\cdot\dfrac{(a^4+1)-(a^4-1)}{(a^4-1)(a^4+1)}$

$=\dfrac{1}{a+1}+\dfrac{2}{a^2+1}+\dfrac{4}{a^4+1}+\dfrac{4}{a^4-1}-\dfrac{4}{a^4+1}$

$=\dfrac{1}{a+1}+\dfrac{2}{a^2+1}+\dfrac{4}{a^4-1}$

$=\dfrac{1}{a+1}+\dfrac{2}{a^2+1}+2\cdot\dfrac{2}{a^4-1}$

$=\dfrac{1}{a+1}+\dfrac{2}{a^2+1}+2\cdot\dfrac{(a^2+1)-(a^2-1)}{(a^2-1)(a^2+1)}$

$=\dfrac{1}{a+1}+\dfrac{2}{a^2+1}+\dfrac{2}{a^2-1}-\dfrac{2}{a^2+1}$

$=\dfrac{1}{a+1}+\dfrac{2}{a^2-1}$

$=\dfrac{1}{a+1}+\dfrac{1+1}{(a-1)(a+1)}$

$=\dfrac{1}{a+1}+\dfrac{(a+1)-(a-1)}{(a-1)(a+1)}$

$=\dfrac{1}{a+1}+\dfrac{1}{a-1}-\dfrac{1}{a+1}$

$=\dfrac{1}{a-1}$