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I couldn't understand how I can get the following result on expansion:

$$3(1-3x)^{-2}=\sum_{n=0}^{\infty}(n+1)3^nx^n.$$ Any ideas?

amWhy
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beginnercoder
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    In the form you have presently, the result is simply false: When $x=0$, the LHS is 3 but the RHS is 1. (The issue is the leading factor of 3 on the LHS, which shouldn't be there.) – Semiclassical Aug 26 '16 at 19:42

1 Answers1

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$$\frac1{(1-3x)^2} =\frac13 \frac{d}{dx} \frac1{1-3x} = \frac13 \frac{d}{dx} \sum_{n=0}^{\infty} (3x)^n \\ = \sum_{n=0}^{\infty} \frac{d}{dx}3^{n-1} x^n = \sum_{n=1}^{\infty}3^{n-1}n x^{n-1} = \sum_{n=0}^{\infty} 3^n (n+1)x^n$$