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What is the expansion for $(1-x)^{-n}$?
Could find only the expansion upto the power of $-3$. Is there some general formula?

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    $\begingroup$ What about the generalized binomial theorem ? $\endgroup$ Jan 24, 2016 at 14:46

6 Answers 6

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I realize this is an old thread, but I wanted to expand on the above answers on how to derive the formula for anyone else that might come along. Starting with the geometric series and taking successive derivatives:

$$ \begin{align} \dfrac{1}{(1-x)} &= 1+x+x^2+x^3+x^4+x^5\dots+x^m+\dotsm\\ \dfrac{1}{(1-x)^2} &= 1+2x+3x^2+4x^3+5x^4\dots+mx^{m-1}+\dotsm\\ \dfrac{2\cdot 1}{(1-x)^3} &= 2+(3\cdot 2)x+(4\cdot 3)x^2+(5\cdot 4)x^3\dots+(m \cdot m-1)x^{m-2}+\dotsm\\ \vdots\\ \dfrac{(n-1)!}{(1-x)^n} &= \sum_{k=0}^\infty \dfrac{(k+n-1)!}{k!} x^k\\ \end{align} $$

which can be simplified by dividing:

$$ \dfrac{1}{(1-x)^n} = \sum_{k=0}^\infty \dfrac{(k+n-1)!}{(n-1)!k!} x^k = \sum_{k=0}^\infty \binom{k+n-1}{n-1}x^k\\ $$

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Yes its the binomial expansion for any index.

$(1-x)^{-n} = (-x)^{0} + -n(-x)^{1}+ \dfrac{-n(-n-1)}{2!}(-x)^{2} + ...$

which simplifies to ..

$(1-x)^{-n} = 1 + nx+ \dfrac{n(n+1)}{2!}(x)^{2} + \dfrac{n(n+1)(n+2)}{3!}(x)^{3} ...$

ie,

$(1-x)^{-n} = 1 + nx+ {n+1\choose 2}(x)^{2} + {n+2\choose 3}(x)^{3} ...$

Binomial expansion for any index is generalization of binomial theorem for positive integral index:

$$(1+x)^n = {n\choose 0} + {n\choose 1}x + {n\choose 2}x^2 + ...$$

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There's a simpler version of the above formula:

$$\frac{1}{(1-x)^n}=\sum_{k=0}^\infty \binom{k+n-1}{n-1}x^k$$

You can prove this by induction - differentiate and then divide by $n$.

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As $\;\biggl(\dfrac1{1-x}\biggr)^{(n-1)}=\dfrac{(n-1)!}{(1-x)^n},\;$ you can derive term by term the power series expansion of $\;\dfrac1{1-x}=1+x+x^2+\dots+x^m+\dotsm$

You obtain \begin{align*} \frac{(n-1)!}{(1-x)^n}&=\sum_{m\ge n-1}m(m-1)\dotsm(m-n+2)x^{m-n+1}\\ &=\sum_{m\ge n-1}\frac{m!}{(m-n+1)!}x^{m-n+1} =\sum_{m\ge 0}\frac{(m+n-1)!}{m!}x^m,\\ \text{whence}\qquad \frac1{(1-x)^n}&=\sum_{m\ge 0}\binom{m+n-1}{m}x^m. \end{align*}

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Here is an (intuitive) proof that explains the coeffcients in $$\sum_{k=0}^\infty \binom{k+n-1 }{n-1} x^k = \frac{1}{(1-x)^n}.$$ Since $$\frac{1}{1-x} = \sum_{k=0}^\infty x^k,$$ we just need to figure out how to multiply it $n$ times. To calculate the coefficient of a specific power, say $x^k$, you only need the first $k+1$ terms in each series. So we have, $$(1+x+x^2+x^3+\cdots +x^k)\cdots (1+x+x^2+x^3+\cdots +x^k).$$

From each multiplicand you need to pick out a power of $x$, and the powers has to be non-negative. These $n$ powers you pick sum up to $k$. So it is the same as asking: How many non-negative integer solutions are there to $$d_1+d_2+\cdots d_n = k.$$

This is given by the "stars and bars" formula in combinatorics, which is exactly $\binom{k+n-1}{n-1}$. For a proof check out the wikipedia page https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

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I wrote this answer up to reduce my own feelings of uncertainty about the expansion.

enter image description here

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    $\begingroup$ @Especially Lime Thank you good Sir. I am not even sure what two characters you deleted to render the image. But thanks! $\endgroup$
    – Arthur W
    Dec 4, 2021 at 2:47

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