High Summation!

Let $n = 1000$ and $p = 0.05$

$E = p^n\sum_{r=0}^{\infty} (n + r) × {{n+r} \choose n} × (1-p)^r$

$\boxed{E = \frac{p^n}{n!} \sum_{r=0}^{\infty} (n + r) × (r+1)(r+2)\ldots(r+n)× (1-p)^r}$

Let's evaluate above sum.

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Let,

$S = \sum_{r=0}^{\infty} (r+1)(r+2)(r+3)\ldots(r+n) ×x^r$

$xS = \sum_{r=0}^{\infty} (r+1)(r+2)(r+3)\ldots(r+n) ×x^{r+1}$

$S(1-x) = \sum_{r=0}^{\infty} n×(r+1)(r+2)(r+3)\ldots(r+n-1) ×x^r$

$xS(1-x) = \sum_{r=0}^{\infty} n×(r+1)(r+2)(r+3)\ldots(r+n-1) ×x^{r+1}$

$S(1-x)^2 = \sum_{r=0}^{\infty} n(n-1)×(r+1)(r+2)(r+3)\ldots(r+n-2) ×x^r$

Continuing, we get

$S(1-x)^n = \sum_{r=0}^{\infty} n×(n-1)×(n-2)×\ldots×2×1 ×x^r = n! \sum_{r=0}^{\infty}x^r$

$\boxed{S = \frac{n!}{(1-x)^{n+1}}}$

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Let, $k \in R^-$ and

$f(k) = \sum_{r=0}^{\infty}e^{k(n+r)}×(r+1)(r+2)\ldots(r+n)×x^r$

$f(k) = e^{kn}\sum_{r=0}^{\infty}(r+1)(r+2)\ldots(r+n)×x^re^{kr}$

Comparing with $S$ , we get

$f(k) = \sum_{r=0}^{\infty}e^{k(n+r)}×(r+1)(r+2)\ldots(r+n)×x^r = \frac{e^{kn}n!}{(1-xe^k)^{n+1}}$

$f'(k) = \sum_{r=0}^{\infty}e^{k(n+r)}×(n+r)×(r+1)(r+2)\ldots(r+n)×x^r = \frac{ne^{kn}(1-xe^{k}) + e^{kn}(n+1)×e^kx}{(1-xe^k)^{n+2}}n!$

Now , $x = 1-p$

$\boxed{f'(0) = \sum_{r=0}^{\infty}(n+r)×(r+1)(r+2)\ldots(r+n)×x^r = n!\frac{(n+1)-p}{p^{n+2}}}$

$E = \frac{p^n}{n!}f'(0)$

$\color{#3D99F6}{\boxed{E = \frac{(n+1)-p}{p^2}}}$

Putting $n = 1000$ and $p = 0.05$ , we get

$\boxed{E=400380}$

We start with the geometric series, $\sum \limits_{r=0}^\infty p^r = \dfrac1{1-p}$ , where $|p| < 1$ . By diffferentiating this equation with respect to $p$ for a total of $n$ times, we will get $\sum \limits_{r=0}^\infty \binom{r+n}n p^r = \dfrac1{(1-p)^{n+1}} .$ Hence, $\newcommand{\sumsum}{\displaystyle \sum_{r=0}^\infty} \newcommand{\bignewline}{ \\ \phantom0 \\ } \begin{array} {r c l} \sumsum (r + n) \binom{r+n}n p^r &=& \sumsum (r + n + 1 - 1) \binom{r+n}n p^r \bignewline &=& \sumsum (r + n + 1) \binom{r+n}n p^r - \sumsum \binom{r+n}n p^r \bignewline &=& \dfrac{d}{dp} \left[ \sumsum \binom{r+n}n p^r \right] - \left[ \sumsum \binom{r+n}n p^r \right] \bignewline &=& \dfrac{n+1}{(1-p)^{n+2}} - \dfrac 1{(1-p)^{n+1}} \bignewline &=& \dfrac{p + n}{(1-p)^{n+2}} \end{array}$ In this case, $n=1000, p = 0.95$ , $\sum_{r=0}^\infty (r + 1000) \binom{r+1000}{1000} \cdot 0.95^r = \dfrac{0.95 + 1000}{0.05^{1002}} \quad \Leftrightarrow \quad 0.05^{1000} \sum_{r=0}^\infty (r + 1000) \binom{r+1000}{1000} \cdot 0.95^r = \dfrac{1000.95}{0.05^2} = \boxed{400380}.$