2010/012) If 0 < x < 1, then the sum of the infinite series (1/2)X^2 + (2/3)X^3 + (3/4)X^4 + ... is ...

A) log[(1+x)/(1-x)];
B) x/(1-x) + log(1+x);
C) 1/(1-x) + log(1-x);
D) x/(1-x) + log(1-x);
Explain your answer...

et f(x) = (1/2)x^2 + (2/3)x^3 + (3/4)x^4 +

df/dx = x + 2 x^2 + 3x^3 + ... ....
= (x+x^2 + x^3...) + (x^2 + x^3 ...) + (x^3+x^4 + ....)
= x/(1-x) + x^2/(1-x) +
= x/(1-x)^2
= (1-x-1)/(1-x)^2
= 1/(1-x) - 1/(1-x)^2
integrating you get
f = log(1-x) + 1/(1-x) + constant of integration log(1-x) as 1- x > 0
this can be checked to be zero so ans is C