How is $\dfrac1{(1-x)^5}=\sum_{n\geq0}{n+4\choose4}x^n$ Can someone please explain to me how is $$\dfrac1{(1-x)^5}=\sum_{n\geq0}{n+4\choose4}x^n$$
Thanks!
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$\ds{\pars{1 - x}^{-5} = \sum_{n = 0}^{\infty}{-5 \choose n}\pars{-1}^{n}x^{n}}$. However ( see expression $\pars{5}$
here ):
$$
{-5 \choose n} = \pars{-1}^{n}{-\bracks{-5} + n - 1 \choose n}
=
\pars{-1}^{n}{n + 4 \choose n} 
=
\pars{-1}^{n}{n + 4 \choose 4} 
$$
$$
\mbox{Then,}\quad\pars{1 - x}^{-5} = \sum_{n = 0}^{\infty}{n + 4 \choose 4}x^{n}
$$
A: Firstly$$\dfrac1{(1-x)^5}=(1-x)^{-5}$$
then from binomial theorem
$$(1-x)^{-5}=\sum_{n\geq0}(-1)^n\binom{-5}{n}x^n$$
because
$$\binom{-5}{n}=\frac{(-5)(-5-1)(-5-2)...(-5-(n-1))}{n(n-1)(n-2)..1}=$$
$$=(-1)^n\frac{(5)(5+1)(5+2)...(5+(n-1))}{n(n-1)(n-2)..1}=$$
$$=(-1)^n\frac{(n+4)(n+3)(n+2)...6\cdot5}{n(n-1)(n-2)..1}=$$
$$=(-1)^n\binom{n+4}{n}=(-1)^n\binom{n+4}{4}$$
finally
$$\dfrac1{(1-x)^5}=\sum_{n\geq0}(-1)^n\binom{-5}{n}x^n=\sum_{n\geq0}\binom{n+4}{4}x^n$$
A: If you like a low-level approach, you can just take for any fixed $k\in\Bbb N$ the formal series $S(k)=\sum_{n\geq0}\binom{n+k}kX^n$. One has $S(0)=\frac1{1-X}$ by the formula for a geometric series. Otherwise
$$
 \begin{align}
  (1-X)S(k)&=
  1+\sum_{n>0}(\binom{n+k}k-\binom{n-1+k}k)X^n
  \\&=\sum_{n\geq0}\binom{n+k-1}{k-1}X^n
 =S(k-1).
 \end{align}
$$
It follows by induction on $k$ that $S(k)=(1-X)^{-k-1}$ for all$~k$. Then put $k=4$.
A: Let $k$ be an integer, and let the sequence $$h_0,h_1,h_2,...,h_n,...$$ be defined by letting $h_n$ equal the number of non-negative integral solutions of $$e_1+e_2+\cdots+e_k=n.$$
We know that the number of non-negative integral solutions to the above equation is $$h_n={n+k-1\choose n}$$ where $n\ge0$. The generating function $$G(x)=\sum_{i=0}^\infty {n+k-1\choose n}x^n={1\over (1-x)^k}.$$
Consider $${1\over (1-x)^k}={1\over 1-x}\cdot{1\over 1-x}\cdots{1\over 1-x}$$ with ($k$ factors). Using the geometric series we obtain $$(1+x+x^2+\cdots)(1+x+x^2+\cdots)\cdots(1+x+x^2+\cdots).$$ Which is equivalent to $$(\sum_{e_1=0}^\infty x^{e_1})(\sum_{e_2=0}^\infty x^{e_2})\cdots(\sum_{e_k}^\infty x^{e_k}).$$ Each $e^i$ is a typical $i$ factor and multiplying all these terms we obtain $$x^{e_1}\cdot x^{e_2}\cdots x^{e_k}=x^n$$ provided that $$e_1+e_2+\cdots+e_k=n.$$ So, we can see that the coefficient of $x^n$ is equal to the number of non-negative integral solutions of the above equation or $${n+k-1\choose n}.$$
Thus $$\sum_{n=0}^\infty {n+k-1\choose n}x^n={1\over (1-x)^k}.$$ Now let $k=5$ and we obtain $${1\over (1-x)^5}=\sum_{n=0}^\infty {n+4\choose n}x^n.$$
A: Here is an approach, which surprisingly, no one has mentioned so far. Using geometric series, we have
$$\dfrac1{1-x} = \sum_{n=0}^{\infty} x^n$$
Differentiating it $k$ times gives us
$$\dfrac{d^k}{dx^k}\left(\dfrac1{1-x} \right) = \sum_{n=k}^{\infty} n(n-1) \cdots(n-k+1)x^{n-k} = \sum_{n=0}^{\infty} (n+k)(n+k-1)\cdots (n+1)x^{n}$$
The left hand side is
$$\dfrac{d^k}{dx^k}\left(\dfrac1{1-x} \right) = \dfrac{k!}{(1-x)^{k+1}}$$
We hence have
$$\dfrac1{(1-x)^{k+1}} = \sum_{n=0}^{\infty} \dbinom{n+k}k x^n$$
A: 
Now for $|x|<1$ suppose that $A=1+x+x^2+x^3+\cdots$
Calculate 4th derivitative of $A$.
A: The coefficient of $x^n$ in $\frac{1}{(1-x)^5}$ counts the number of partitions of $n$ into five nonnegative integers.
Consider a $(0,1)$-sequence of length $n+4$ consisting of $n$ ones and four zeroes.
Such sequences are equivalent to partitions of $n$ into five parts and thus
$$ \frac{1}{(1-x)^5} = \sum \binom{n+4}{4} x^n $$.
