First, convert the problem to a slightly different form: \(\frac{1}{(a+b)^n} = \frac{1}{b^n} \frac{1}{(\frac{a}{b}+1)^n}
= \frac{1}{b^n} \sum_{k=0}^{\infty} {(-1)^k \binom{n + k - 1}{k} \left ( \frac{a}{b} \right ) ^k}\)
So, let’s replace \(\frac{a}{b} = x\) and ignore for a while the term factored out. Then, we only need to show
\begin{equation*}
\sum_{k=0}^{\infty} {(-1)^k \binom{n + k - 1}{k} x^k} = \left ( \frac{1}{1+x} \right )^n
\end{equation*}
However
\begin{align*}
\left ( \frac{1}{1+x} \right )^n & = \left ( \frac{1}{1 - (-x)} \right )^n \\
& = \left ( \sum_{k=0}^{\infty} {(-1)^k x^k} \right )^n
\end{align*}
This infinite sum raised to a power can be expanded by distributing terms in the standard way. In doing so, the various powers of x multiplied together will create a series in powers of x involving \(x^0, x^1, x^2, ...\text{.}\) To detemine the final coefficients notice that the number of time \(m^k\) will appear in this product depends upon the number of ways one can write k as a sum of nonnegative integers.
For example, the coefficient of \(x^3\) will come from the n ways of multiplying the coefficients \(x^3, x^0, ..., x^0\) and \(x^2, x^1, x^0, ..., x^0\) and \(x^1, x^1, x^1, x^0,..., x^0\text{.}\) This is equivalent to finding the number of ways to write the number k as a sum of nonnegative integers. The possible set of nonnegative integers is {0,1,2,...,k} and one way to count the combinations is to separate k *’s by n-1 |’s. For example, if k = 3 then *|*|* means \(x^1 x^0 x^2 = x^3\text{.}\) Similarly for k = 5 and |**|*|**| implies \(x^0 x^2 x^1 x^2 x^0 = x^5\text{.}\) The number of ways to interchange the identical *’s among the idential |’s is \(\binom{n+k-1}{k}\text{.}\)
Furthermore, to obtain an even power of x will require an even number of odd powers and an odd power of x will require an odd number of odd powers. So, the coefficient of the odd terms stays odd and the coefficient of the even terms remains even. Therefore,
\begin{equation*}
\left ( \frac{1}{1+x} \right )^n = \sum_{k=0}^{\infty} {(-1)^k \binom{n + k - 1}{k} x^k}
\end{equation*}
Similarly,
\begin{equation*}
\left ( \frac{1}{1-x} \right )^n = \left ( \sum_{k=0}^{\infty} {x^k} \right )^n = \sum_{k=0}^{\infty} {\binom{n + k - 1}{k} x^k}
\end{equation*}
To validate the above proof using software, consider the interactive cell below that determines each side of the above formula and shows that they are equal.