Given
$$(1-\frac{x}{10})^{-3}=1+\frac{3}{10}x+\frac{3}{50}x^2+\frac{1}{100}x^3+\frac{3}{2000}x^4+\cdots$$
Show that the coefficient of $x^n$ in the expansion of $(1-\frac{x}{10})^{-3}$ is $\frac{1}{2}(n+1)(n+2)\frac{1}{10^n}$.
After evaluating a few terms, it is evident the statement is true. However, I am struggling to prove this rigorously. Induction seems unnecessarily complicated so I think a proof by deduction would be best where you prove for a general coefficient in the binomial expansion of $(1-\frac{x}{10})^{-3}$, though the cancellations become quite confusing with so many negatives etc..
Thanks
I have used the formula
$(1+x)^n = 1 +nx + \frac{n(n-1)}{2!}x^2 +\frac{n(n-1)(n-2)}{3!}x^3+...+$
which gives for $x^n$ coefficient in the binomial expansion of $(1-\frac{x}{10})^{-3}$
$\frac{(-3)(-4)(-5)...(-3-(n-2))(-3-(n-1))}{n(n-1)(n-2)...3\times2\times1}(-\frac{1}{10}x)^n)$
I am not sure how to simplify this properly, but I want to specifically used this formula??
Given
$(1+x)^k = 1 +kx + \frac{k(k-1)}{2!}x^2 +\frac{k(k-1)(k-2)}{3!}x^3+...+$
$= \sum_{n=0}^\infty \frac{k(k-1)(k-2)...(k-n+1)}{n!}x^n$
$=\sum_{n=0}^\infty \frac{k(k-1)(k-2)...(k-n+1)(k-n)(k-n-1)...2\times1}{n!(k-n)(k-n-1)...2\times1}x^n$
$=\sum_{n=0}^\infty \frac{k!}{n!(k-n)!}x^n$
$=\sum_{n=0}^\infty {k \choose n}x^n$
Which links the two general results together.
Therefore the coefficient of $x^n$ for the expansion of $(1-\frac{x}{10})^{-3}$ is
${-3\choose n}$
The generalized binomial theorem gives $$(1 - \frac{x}{10})^{-3} = \sum_{k = 0}^\infty \binom{3 +n-1}{n} \frac{1}{10^n} x^n = \sum_{k = 0}^\infty \binom{n+2}{2}\frac{1}{10^n} x^n\ $$ which is exactly what you have.
EDIT: Here's a calculation \begin{align*} \binom{-3}{n}(-1)^n &= \frac{(-3)(-4)\cdots(-3 - (n-1))}{n!}(-1)^n \\ &=\frac{3\cdot 4 \cdots (n+2)}{n!} \\ &=\frac{(n+2)(n+1)}{2} \end{align*}