Question

Guys I am preparing for jee advance my exam is on 21 may can u help in in this question
1/(1+x) + 2x/1+x^2 + 4x^3/1+x^4 +........2^n.x^(2n-1)/1+x^2^n
Help me in this question

Answer 1

1/x+1 + 2x/1+x^2  + 4x^3/1+x^4.................................2^2nx^2n-1/1+x^2n
enrique just follow my steps what i am doing ^_^ 
the upper part is differentiation of the lower part just like in the format of
f'(x)/1+f(x) = function
so i am using the concept of limits and derivatives to solve your question
let
Y = 1/x+1 + 2x/1+x^2  + 4x^3/1+x^4.................................2^2nx^2n-1/1+x^2n
integration of 1/x = lnx now integration of 1/1+x = ln(1+x) using this property i will solve your question
∫y= ∫1/1+x  + ∫2x/1+x^2+ ....................................................∫2x^2n-1dx/1+x^2n
∫y= ln(1+x)+ln(1+x^2)................................................ln(1+x^2n)
now using property of log they are at same base so we can directly multiply all
∫y=ln[(1+x)(1+x^2)............................(1+x^2^2n)]
now i will multiply and divide (1-x) in the logarithmic equation
∫y= ln[(1-x)(1+x)(1+x^2)...................(1+x^2^2n)]/(1-x)
∫y=ln[(1-x^2)(1+x^2).........................(1+x^2^2n)/(1-x)
similarly it is breaking down in the form of a formula
so finally we get -;
∫y=ln(1-x^4^2n) - ln (1-x) +c
now differentiate the whole term wrt x cause we have integrated it now for making a balance  i am just differentiating it
differentiation of ln x = 1/x
y = 1/(1-x^4^2n)  -  1/(1-x)
when n tends to infinite 
1/(1-x^4^2n) tends to zero
y= 1/(x-1) is the sum of the series hope you got it

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