Sum of $k-tuple$ partition of $d$. $k$ and $d$ are both positive integers.

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I am trying to understand this paper by Bondy and Jackson and in it I found the following calculation in which I cannot figure out how we got from 2nd last step to last step. I have referred the paper for the further details of this calculation.

I understand that in the initial steps were are using power series after that we are manipulating the terms using partial fraction.

Here $c(k,d)$ is the cardinality of ordered $k$ tuple summing to $d$ and $2^{k-1}$ is weight of those tuples.

$\sum _{k\geq 2} 2^{k-1} c( k,d) \ \\ \\ \vdots \\ =\ \left[ x^{d}\right]\frac{1}{3}\left(\frac{2}{1-2x} -\frac{3}{1-x} +\frac{1}{1+x}\right)\\ \\ =\ \frac{1}{3}\left( 2^{d+1} -3+( -1)^{d}\right) \blacksquare $

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Note that $[x^d]$ denotes the coefficient of $x^d$ in a series. We obtain by geometric series expansion: \begin{align*} \color{blue}{[x^d]\frac{2}{1-2x}}=2[x^d]\sum_{n=0}^\infty (2x)^n=2\cdot 2^d\color{blue}{=2^{d+1}} \end{align*} and similarly for the other terms.