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Find the coefficient of $x^{27}$ in the sequence generated by the function $\lambda x\in\mathbb{R}.$ ${b^mx^m}\over(1-bx)^{m+1}$ where $b\in\mathbb{R^{+}}$ and $m\in\mathbb{N}$ are some parameters.

Hello all. I need to find the coefficient of $x^{27}$. I know the function $h(x)=(1-bx)^{-m-1}$ generates the sequence $\lambda n\in\mathbb{N}.\binom{m+1+k-1}{m+1-1}$

So overall if $m>27$ then $a_{27}=0$, else, I get $a_n=b^{27}\binom{m+27}{m}$, I know I'm doing something wrong because the correct answer is $b^{27}\binom{27}{m}$.

Would love to get your help on that one, thank you!

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    $\begingroup$ You forgot the factor $b^mx^m$, your answer being correct for $1/(1-bx)^{m+1}$. $\endgroup$
    – Did
    Jan 13, 2018 at 12:17

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$$\frac{1}{(1-y)^{m+1}} = \sum_{n \geq 0} \binom{n}{m}y^{n-m}$$

Where you can notice this pattern by differentiating $\frac{1}{1-y} = \sum_{n\geq 0} y^n$ multiple times.

Thus: $$ \begin{align} f(x)=\frac{(bx)^m}{(1-bx)^{m+1}} &= (bx)^m \sum_{n \geq 0} \binom{n}{m}(bx)^{n-m} \\ &= \sum_{n \geq 0} \binom{n}{m}(bx)^n \end{align}$$

So $[x^{27}]f(x) = \binom{27}{m}b^{27}$, which equals $0$ if $m > 27$.

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For $m\lt 27$ let the coefficient of $x^r$ in the expansion of $ (1-bx)^{-m-1}$ along with $x^m$ outside the expansion give coefficient of $x^{27}$. Hence we get $r=27-m$.

Hence the coefficient of $x^{27}$ in given expression would be $$b^{27} \binom {m+1+27-m-1}{27-m} = b^{27} \binom {27}{m}$$

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