Approximate a series with finite number of terms How it is possible to approximate:
$$\sum_{i=1}^{NR}{i\cdot \left( \dfrac{1}{1-p} \right)^i} $$

Let $x = \dfrac1{1-p}$ and let $n = NR$. Then we are interested in the sum $\displaystyle \sum_{i=1}^{n} i x^i$.
\begin{align}
\sum_{i=1}^{n} i x^i & = x \left(\sum_{i=1}^{n} i x^{i-1} \right)\\
& = x \left( \sum_{i=1}^{n} \frac{d x^i}{dx} \right)\\
& = x \frac{d}{dx} \left( \sum_{i=1}^{n} x^i\right)\\
& = x \frac{d}{dx} \left( x\left(\frac{x^n - 1}{x-1} \right) \right)\\
& = x \left( \frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2} \right)
\end{align}
Replacing $x$ by $\dfrac1{1-p}$ and $n$ by $NR$, we get that the solution is $$\left( \frac{NR}{p} - \frac1{p^2} + \frac1p \right) \left( \dfrac1{1-p}\right)^{NR} + \frac1{p^2} - \frac1p$$
