Calculate the binomial $(1-x)^{-(n+1)}$

Calculate the binomial $(1-x)^{-(n+1)}$

What I have tried:

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

However, I was wondering how I could get this form instead: $$(1-x)^{-(n+1)} = \sum_{n=k}^{\infty} \binom{n}{k}x^{n-k}?$$

Solution 1:

It is more convenient to take $n+1$ as is which gives \begin{align*} \color{blue}{(1-x)^{-(n+1)}}&=\sum_{k=0}^{\infty}\binom{-(n+1)}{k}(-x)^k\tag{1}\\ &=\sum_{k=0}^\infty\binom{n+k}{k}x^k\tag{2}\\ &=\sum_{n=0}^\infty\binom{k+n}{n}x^n\tag{3}\\ &=\sum_{n=k}^\infty\binom{n}{n-k}x^{n-k}\tag{4}\\ &\,\,\color{blue}{=\sum_{n=k}^\infty\binom{n}{k}x^{n-k}}\tag{5} \end{align*} and the claim follows.

Comment:

• In (1) we use the binomial series expansion.

• In (2) we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (3) we exchange $k$ and $n$.

• In (4) we shift the index to start with $n=k$.

• In (5) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.