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Use the two formulas $\sum_{n=0}^{\infty}=\frac{1}{1-x}$ and $\sum_{n=0}^{\infty}\left ( n+1 \right )x^{n}=\frac{1}{(1-x)^{2}}$ to find a formula for this $\sum_{n=0}^{\infty}\left ( an+b \right )x^{n}$ for all pair of constants $a$ and $b$ (both non-zero).

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Hint: $an+b = a(n+1) + (b-a)$ – Thomas Andrews May 31 '13 at 19:28
AjmalW: FYI, the question was most likely downvoted because it is stated as a command, and does not include the your thoughts on the problem. It is best to show what effort you've put in so far, to indicate that you're not just looking for people to do your homework as well as to avoid getting answers that tell you what you already know. – Stahl May 31 '13 at 19:35
Stahl, I have tried to do on my own by differentating and other things but I failed. I didn't intend to ask here what the answer (result) is otherwise I would fail in the exam. When I read other answers (hint) it made me feel relieved to know where to start. Thanks for your comment. – AjmalW Jun 1 '13 at 10:19
up vote 5 down vote accepted



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

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Why the downvote? This contains all the useful information ($\sum nx^n$ and $\sum x^n$) in one single formula. Pretty nice. Hmm, I see, no use of $\sum (n+1)x^n$... Well, does it really matter? – 1015 May 31 '13 at 19:33
Thanks @julien. Likely someone interpreted the question as HW (even though that is far from obvious) and was not appreciative of my essentially delivering the answer. But I am guessing. Oh, and the $(n+1) x^n$ is replaced by the factor of $x$ outside. – Ron Gordon May 31 '13 at 19:36
All you've shown is a variant of the given fact that $\sum (n+1)x^n = 1/(1-x)^2$. He already knows this, so showing him the derivative trick doesn't help and really doesn't get him anywhere. – Thomas Andrews May 31 '13 at 20:52

Let us look at (i) $\sum_{n=0}^\infty (n+1)x^n$, which we have been kindly supplied a formula for, and (ii) $\sum_{n=0}nx^n$, which we need.

The first is $$1+2x+3x^2+4x^3+\cdots.\tag{i}$$

The second is $$(0)(1)+x+2x^2+3x^3+4x^4+\cdots,$$ or equivalently $$x+2x^2+3x^3+4x^4+\cdots.\tag{ii}$$

It is clear that (ii) is $x$ times (i), so (ii) is $\dfrac{x}{(1-x)^2}$.

Remark: Note that in the convergence sense, the formulas are only valid if $|x|\lt 1$.

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