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Why $$\sum_{n=0}^{\infty}(n+1)5^nx^n=\frac{1}{(1-5x)^2}?$$

I know that $\sum_{n=0}^{\infty}x^n=\dfrac{1}{1-x}$, so by the same token, $\sum_{n=0}^{\infty}5^nx^n=\dfrac{1}{1-5x}$.

Thus $$ \left(\frac{1}{1-5x}\right)^2=\frac{1}{(1-5x)^2} = \left(\sum_{n=0}^{\infty}5^nx^n\right)^2. $$

But why is $\big(\sum_{n=0}^{\infty}5^nx^n\big)^2=\sum_{n=0}^{\infty}(n+1)5^nx^n$?

Assuming $x$ is small enough so that the sum converges.

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    $\begingroup$ See this. Note your series can be written as $\sum_{n=0}^\infty (n+1)(5x)^n$. $\endgroup$ – David Mitra Feb 5 '14 at 15:56
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Note that $$ \frac{1}{1-x}=\sum_{n=0}^\infty x^n, $$ and thus $$ \frac{1}{(1-x)^2}=\left(\frac{1}{1-x}\right)'=\sum_{n=1}^\infty nx^{n-1}=\sum_{n=0}^\infty (n+1)x^{n}, $$ and hence $$ \frac{1}{(1-5x)^2}=\sum_{n=0}^\infty (n+1)(5x)^{n}. $$

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What happens when you differentiate $\frac{1}{1-5x}$? How does this relate to your series?

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  • $\begingroup$ I see now! Thank you. $\endgroup$ – Oria Gruber Feb 5 '14 at 15:58
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Although the method via differentiating the known power series is what I'd recommend, here's a direct argument why $(\sum_{n=0}^{\infty}5^nx^n)^2=\sum_{n=0}^{\infty}(n+1)5^nx^n$ or more simply why $$(\sum_{k=0}^{\infty}x^k)^2=\sum_{n=0}^{\infty}(n+1)x^n.$$

The left-hand-side is $$(x^0 + x^1 + x^2 + \cdots + x^k + \cdots) \times (x^0 + x^1 + x^2 + \cdots + x^l + \cdots),$$ so the coefficient of $x^n$ in the LHS is the number of pairs $(k, l)$ with $k + l = n$. This is $n+1$, as there are $n+1$ pairs $(0,n), (1,n-1), \dots, (n,0)$. This is the same as the coefficient of $x^n$ in the RHS.

We can also give it a combinatorial interpretation: $F(x) = \sum_{n=0}^{\infty} x^n$ is the generating function for the nonnegative integers, and so $F(x)^2$ is the generating function for the number of ways of writing an integer as the sum of two nonnegative integers, which is $n+1$ (by the argument above).

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