The coefficient of $x^{100}$ in the expansion of $(1-x)^{-3}$. Please solve it and tell me the technique so that I can solve it in examination in multiple choice questions.
 A: For a negative integer n, the binomial theorem gives $(x+a)^{-n}=\sum_{k=0}^{\infty} {-n\choose k}x^ka^{-n-k}$.
So in the case of calculating the coefficient for $x^{100}$ in $(1-x)^{-3}$, $k=100,n=3,a=1$.
${-3\choose 100}(-x)^{100}1^{-3-100}=5151x^{100}$.
So the coefficient would be $5151$.
A: This is an application of the generalized binomial theorem, that
$$(x+y)^s = \sum_{k=0}^\infty {s \choose k} x^{s-k}y^k$$
where
$${s \choose k} = \frac{s(s-1)(s-2)\cdots(s-k+1)}{k!}$$
is the generalized binomial coefficient. Hence in this case,
$$(1-x)^{-3} = \sum_{k=0}^\infty {-3 \choose k} (1)^{-3-k}(-x)^{k}$$
and so taking $k = 100$, the coefficient of the $x^{100}$ term will be
$${-3 \choose 100} = \frac{(-3)(-4)\cdots(-102)}{(100)(99)\cdots(1)} = \frac{(102)(101)}{2} = 5151$$
A: We begin by noting that $\frac{1}{1-x}=(1-x)^{-1}=1+x+x^2+x^3+x^4+\dots$
We see then that $(1-x)^{-3}=(1+x+x^2+x^3+\dots)(1+x+x^2+x^3+\dots)(1+x+x^2+x^3+\dots)$
Approaching from a combinatorial perspective, we recognize that to have arrived at the term $x^n$, we will have needed to selected a term from the first parenthesis, the second parenthesis, and the third such that their exponents add up to $n$.  We further recognize that each such way of doing this will contribute one to the overall coefficient and there are no other ways to arrive at that term.
The question then is related to:

How many non-negative integer solutions are there to the system $$\begin{cases}x_1+x_2+x_3=n\\0\leq x_i\end{cases}$$

From earlier example, we know the answer to be $\binom{n+3-1}{3-1}$ via stars and bars.
For the specific example of $n=100$, this would imply the coefficient to be $\binom{102}{2}=5151$
A: Here is a different approach. We have the geometric series $\frac{1}{1-x}=1+x+x^2+x^3+x^4+...$ Now take the derivative twice and divide left hand right hand side by $2$ to arrive at: $\frac{1}{(1-x)^3}=1+3x+6x^2+...$ which in summation form is $\frac{n^2-n}{2}x^{n-2}$. Since the exponent has to be $100$, it follows that $n=102$ from which the coefficient follows.
A: The function:
$$f(x)=\dfrac{1}{(1-x)^3}$$ can be expanded in series as:
$$f(x)\simeq\sum_{k=0}^N\dfrac{1}{2}{x^k}(k+1)(k+2)$$
so the $100^{th}$ coefficient is: $C(100)=\dfrac{1}{2}(101)(102)=5151$
