Answered

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(a) Starting with the geometric series [infinity] xn n = 0 , find the sum of the series [infinity] nxn − 1 n = 1 , |x| < 1.

Sagot :

LammettHash

Let f(x) be the sum of the geometric series,

[tex]f(x)=\displaystyle\frac1{1-x} = \sum_{n=0}^\infty x^n[/tex]

for |x| < 1. Then taking the derivative gives the desired sum,

[tex]f'(x)=\displaystyle\boxed{\dfrac1{(1-x)^2}} = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}[/tex]

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