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$$\frac{x}{{{{(1 - x)}^6}}} = x\sum\limits_n^\infty {\left( {\begin{array}{*{20}{c}} {n + 5} \\ 5 \\ \end{array}} \right)} {x^n} = \sum\limits_n^\infty {\left( {\begin{array}{*{20}{c}} {n + 5} \\ 5 \\ \end{array}} \right)} {x^{n + 1}}$$

The conclusion is that the coefficient of $x^n$ is $\left( {\begin{array}{*{20}{c}} {n + 4} \\ 5 \\ \end{array}} \right)$.

Can you explain where the $x$ (middle expression) came from?
And why the coefficient of $x^n$ is $\left( {\begin{array}{*{20}{c}} {n + 4} \\ 5 \\ \end{array}} \right)$? maybe it's a typo.

AndrePoole
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1 Answers1

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The middle expression is the result of iterated use of the power rule. If you take the $n^\text{th}$ (formal) derivative of $\frac1{1-x}$, you end up with $\frac{1}{(1-x)^{n+1}}$ times some polynomial in $n$. If you then take the fifth derivative of the geometric series, then you can work out the coefficient to be $\binom{n+5}{5}$ but missing some terms; those terms come from that extra polynomial in $n$.

The reason you end up with $n+4$ instead of $n+5$ will be obvious if you are more careful about the indices of summation.

Eric Stucky
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  • Oh, sorry, I didn't mention that. It's because you have an $x$ in the numerator of the first expression. You could maybe be fancy and try to incorporate it into the series somehow but that seems a lot harder than the tactic of "ignore it for now and hope the problem disappears later". Since the problem does disappear later, you're fine :P – Eric Stucky Mar 02 '14 at 02:20