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What is the series representation of $(1+x)^{-n}$, where $n$ is a positive integer? I have this term in an integral, and I want to replace this term by a series representation to be able to solve the integral.

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You may also recall the binomial series expansion \begin{align*} (1+x)^{-n}&=\sum_{k=0}^\infty \binom{-n}{k}x^k\tag{1}\\ &=\sum_{k=0}^\infty \binom{n+k-1}{k}(-1)^kx^k \end{align*}

In (1) we use a binomial identity with negative integers.

Since the definition of a binomial coefficient with general $\alpha\in\mathbb{R}$ is \begin{align*} \binom{\alpha}{k}=\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-(k-1))}{k!} \end{align*} we obtain \begin{align*} \binom{-n}{k}&=\frac{-n(-n-1)(-n-2)\cdots(-n-(k-1))}{k!}\\ &=(-1)^k\cdot\frac{n(n+1)(n+2)\cdots(n+(k-1))}{k!}\\ &=(-1)^k\binom{n+k-1}{k} \end{align*}

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    i reached (1), but i couldn't complete because of the negative sign in (n) – 2017-02-01
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    @user42138: Derivation added (see also the referred link). – 2017-02-01
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    great, thank you so much. – 2017-02-01
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    @user42138: You're welcome! :-) – 2017-02-01
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Recall that $$\frac1{1+x} = \sum_{k=0}^\infty (-1)^k x^k$$

Now take $(n-1)$ derivatives on both sides.

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    (n-1) derivatives? can you show me how this may be done?? Thank you a lot. @J.R. – 2017-02-01
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    Well take the first derivative then you get – 2017-02-01
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    $-1/(1+x)^2 = \sum_{k=1}^\infty (-1)^k k x^{k-1}$ .. etc. – 2017-02-01
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    Second derivative, $2/(1+x)^3 = \sum_{k=2}^\infty (-1)^k k(k-1) x^{k-2}$.. can you come up with a pattern and write down the $(n-1)$th derivative? – 2017-02-01
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    oh i see, is there a general formula for the (n-1) derivatives, and i have a question; why (n-1)? why not n? @J.R. – 2017-02-01
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    Because the $(n-1)$th derivative of $1/(1+x)$ is what you want: constant times $1/(1+x)^n$. The $n$th derivative would leave us with constant times $1/(1+x)^{n+1}$. – 2017-02-01
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    Now you do the rest (i.e. come up with the formula for $n-1$). It's a good exercise, trust me. – 2017-02-01
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    i see, ok i'll do it, after i get a general pattern, is this will be the expansion to my term? @J.R. – 2017-02-01
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    Yes. Then, you will also be able to simplify it and on the right hand side there should appear a certain [binomial coefficient](https://en.wikipedia.org/wiki/Binomial_coefficient) after simplification. – 2017-02-01
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    great. i'll do it now, thank you so much.@J.R. – 2017-02-01
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    done. i got: $\frac{(n-1)! (-1)^{n-1}}{(1+x)^n}=\sum_{k=n-1}(-1)^k \prod_{r=0}^{n-2}(k-r) x^{k-(n-1)}$ @J.R. – 2017-02-01
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    Correct. Now simplify. – 2017-02-01
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    I couldn't simplify it more...what can i do more? – 2017-02-01
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    Put constants on right hand side and index shift in the summation. – 2017-02-01
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    i got this : $\sum_{k=0}^{\infty}{ (-1)^k{{k+n-1}\choose {n-1}} x^k}$, is it right?? – 2017-02-05