Finding the sum of geometric progression Evaluate the sum:
$$\sum_{x=0}^\infty x(x-1) {2+x \choose x}(0.008)(0.8)^x $$
I was able to make this into:
$$0.004\sum_{x=0}^\infty x(x-1) (x+1)(x+2)(0.8)^x $$
Let $x=n-2$ then $n=x+2$:
$$0.004\sum_{n=2}^\infty n(n-1)(n-2)(n-3)(0.8)^{n-2} $$
$$0.004\sum_{n=4}^\infty (n)_3~(0.8)^{n-2} $$
Let $j=n-4$ then $n=j+4$:
$$0.004\sum_{j=0}^\infty (j+4)_3~(0.8)^{j+2} $$
$$0.00256\sum_{j=0}^\infty (j+4)_3~(0.8)^{j} $$
I thought I could just use the formula for geometric sum on this but I know the answer's $192 $ and right now my solution's pretty far off.

$$(x-1) x \binom{x+2}{x}=\frac{1}{2} (x-1) x (x+1) (x+2)$$ Rewriting
$$\frac{1}{2}(x-1) x (x+1) (x+2)= \frac{1}{2}x (x - 1) (x - 2) (x - 3) + 4 x (x - 1) (x - 2) + 6 x (x - 1) $$
$$S=\sum_{x=0}^\infty x(x-1) {2+x \choose x}p^x=\frac{1}{2}p^4\sum_{x=0}^\infty x (x - 1) (x - 2) (x - 3)p^{x-4}+$$ $$4p^3\sum_{x=0}^\infty x (x - 1) (x - 2) p^{x-3}+6p^2\sum_{x=0}^\infty x (x - 1)  p^{x-2}$$
$$S=\frac{1}{2}p^4\Bigg[\sum_{x=0}^\infty p^x\Bigg]''''+4p^3\Bigg[\sum_{x=0}^\infty p^x\Bigg]'''+6p^2\Bigg[\sum_{x=0}^\infty p^x\Bigg]''$$
Since
$$\sum_{x=0}^\infty p^x=\frac 1 {1-p}$$ just compute the second, third and fourth derivatives and get
$$S=\frac{12 p^2}{(1-p)^5}$$ If $p=0.8$ then $S=24000$ which, multiplied by $0.008$ gives $192$.
