<s>[INST] Intuitive explanation of $(1-x)^{-a-1}=\sum_{j=0}^{\infty}{{a+j} \choose j}x^j$ Could anyone please explain me the reasoning behind this formula?
$(1-x)^{-a-1}=\sum_{j=0}^{\infty}{{a+j} \choose j}x^j$
Thanks so much!
 [/INST] From the sum of geometric series, we know that $1/(1-x)=1+x^2+x^3+\cdots$, for $|x|<1$.
Thus, one can perceive $(1-x)^{-a-1}=1/(1-x)^{a+1}$ as the $a+1$th power of $1/(1-x)$, given $a$ is an integer,
$$
\underbrace{\frac{1}{1-x}\cdot \frac{1}{1-x} \cdots \frac{1}{1-x}}_\text{a+1} \\
=\underbrace{(1+x^2+x^3+\cdots)\times(1+x^2+x^3+\cdots)\times \cdots \times (1+x^2+x^3+\cdots)}_\text{a+1} \tag{*}
$$
From $(*)$, we know that the coefficient of $x^j$ is the number of solutions for the integer equation $m_1+m_2+\cdots+m_{a+1}=j$, with $m_k=0,1,\ldots$, which is ${a+j \choose j}$.
Therefore, we can deduce that $(1-x)^{-a-1}=\sum_{j=0}^{\infty}{{a+j} \choose j}x^j$.
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