Finding the sum of geometric progression

Question

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.

Answer 1

There are a couple ways to go about this. An important identity is that $\frac{1}{(1-x)^k}=\sum_{n=0}^\infty \binom{n+k-1}{k-1}x^n$ for $|x|<1$. This can be proven with just binomial theorem and the negative binomial coefficients. It can also be proven by noting that $\frac{1}{(1-x)^k}=(1+x+x^2+\ldots)^k$. The coefficient of $n$ is $\binom{n+k-1}{k-1}$ by stars-and-bars.

Method 1

Let's say we wanted to find the value of $$g(x)=\sum_{n=0}^\infty n(n-1)\binom{2+n}{n} x^n$$ Consider the function $f(x)=\frac{1}{(1-x)^3}=\sum_{n=0}^\infty \binom{n+2}{n}x^n$. We know that $$f''(x)=\frac{(3)(4)}{(1-x)^5}=\sum_{n=0}^\infty n(n-1)\binom{n+2}{n}x^{n-2}$$ So $$g(x)=x^2f''(x)=\frac{12x^2}{(1-x)^5}$$ Our desired expression is the value of $.008g(.8)=\frac{12(.008)(.8^2)}{(.2)^5}=\frac{12(8^3)}{2^5}=\boxed{192}$

Method 2

This aligns a little more closer with your original method. You already showed that the desired expression is equivalent to $$.004\sum_{n=0}^\infty (n+2)(n+1)(n)(n-1)(.8)^n$$ $$=.004(4!)\sum_{n=0}^\infty \binom{n+2}{4}(.8)^n$$ $$=.004(4!)\sum_{n=0}^\infty \binom{n+2}{4}(.8)^n$$ Note that $\binom{n+2}{4}=0$ for $n=0,1$, so this sum is equivalent to $$=.004(4!)\sum_{n=2}^\infty \binom{n+2}{4}(.8)^n$$ $$=.004(4!)\sum_{n=2}^\infty \binom{(n-2)+4}{4}(.8)^n$$ We can shift the indices down by $2$ to see that this is equivalent to $$=.004(4!)\sum_{n=0}^\infty \binom{n+4}{4}(.8)^{n+2}$$ $$=.004(.8)^2(4!)\sum_{n=0}^\infty \binom{n+4}{4}(.8)^n$$ From our initial identity, we know that $\sum_{n=0}^\infty \binom{n+4}{4}(.8)^n=\frac{1}{(.2)^5}$, so our expression is equivalent to $$=\frac{.004(.8)^2(4!)}{(.2)^5}$$ $$=\frac{4(8)^2(4!)}{2^5}$$ $$=\boxed{192}$$

Answer 2

$$(x-1) x \binom{x+2}{x}=\frac{1}{2} (x-1) x (x+1) (x+2)$$ 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 $192$.