Verified answer

Since 1/(1 - x) = sum(n=0 to infinity) x^n for |x| < 1,

differentiating both sides yields

1/(1-x)^2 = sum(n= 1 to infinity) nx^(n-1) = 1 + 2x + 3x^2 + ... .

Note that this holds when |x| < 1.

I hope this helps!

Sum(n*x^n-1)

sum(S)= 1/(1-x)^2

. Let : S = 1 + 2x + 3x² + 4x³ + 5x^4 + ......... (1)

....Then : x.S = x + 2x² + 3x³ + 4x^4 + ..........(2)

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Then, ... Eq(1) - Eq(2) ... gives

S - xS = 1 + x + x² + x³ + ...

S ( 1 - x ) = 1 / ( 1 - x ), ......where ... | x | < 1

S = 1 / ( 1 - x )² ............Ans.

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