Let S = 1/(1-x) + 1/(1+x) + 2/(1+x^2) +4/(1+x^4) + 8/(1+x^6)

First let us calculate the first two terms:

1/(1-x) + 1/(1+x) = 2/(1-x^2

==> S = 2/(1-x^2) + 2/(1+x^2) + 4/(1+x^4) + 8/(1+x^6)

Now we will calculate the first two terms again:

==> 2/(1-x^2) + 2/(1+x^2) = (2+2x^2 + 2(1-x^2)/(1 - x^4)

= 4(1-x^4)

==> S = 4/(1- x^4 ) + 4/(1+x^4) + 8/(1+x^6)

Now calculate first two terms:

4(1-x^4) + 4/(1+x^4) = 4(1+x^4) + 4(1-x^4)]/(1-x^8)= 8/(1-x^8)

**==> S = 8/(1-x^8) + 8/(1+x^6) **

Add the expression succesively from the left:

1/(1-x) +1/(1+x) = (1+x+1-x)/(1-x)(1+x) = 2/(1-x^2)

2/(1-x^2) +2/(1+x^2) = 2(1+x^2+1-x^2)/(1-x^2)(1+x^2) = 4/(1-x^4)

4/(1-x^4) +4/(1+x^4) = 4 (1+x^4+1-x^4)/(1-x^4)(1+x^4) = 8/(1-x^8.). Now add the last term 8/(1+x^8):

8/(1-x^8) +8/(1+x^8) = 8(1+x^8 +1-x^8)/(1-x^8)(1+x^8)

= 16/(1-x^16)

So the sum of the terms = 16/(1-x^16)

We'll combine and eliminate the like terms from numerator and we'll have:

1/(1-x)+1/(1+x) = 2/(1-x^2)

The new quotient will be added to the third term.

[2/(1-x^2)]+ [2/(1+x^2)]

We'll have as LCD , the difference of squares:

LCD = (1-x^2)(1+x^2) = 1-x^4

[2/(1-x^2)]+ [2/(1+x^2)]=[2(1-x^2)+2(1+x^2)]/( 1-x^2)( 1+x^2)

We'll remove the brackets and we'll combine and eliminate like terms:

[2/(1-x^2)]+ [2/(1+x^2)] = 4/(1-x^4)

The new quotient will be added to the fourth term.

4/(1-x^4) + 4/(1+x^4)

Judging according to the the previous results, we'll get:

4/(1-x^4) + 4/(1+x^4) = 8/(1-x^8)

The new quotient will be added to the fifth term.

8/(1-x^8)+ 8/(1+x^8)=16/(1-x^16)

The final result of the sum is:

**S = 16/(1-x^16)**