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.