Closed for solution for $\sum_{k = 0}^{n} Q^{k} ( 1 - Q) ^ {k}$

Question

I know the binomial expansion formula:

$$ (1 + x)^n = \sum_{k = 0}^{n} {n \choose k}x^k $$ However, I am trying to find (if there is any) a closed-form solution for the following equation. $$ \sum_{k = 0}^{n} Q^{k} ( 1 - Q) ^ {k} $$ Could you guys point me to some solutions? TIA.

Answer 1

Use the geometric series $\sum_{k=0}^nr^k=\frac{1-r^{n+1}}{1-r}$ with $r=Q(1-Q)$ to get $\frac{1-Q^{n+1}(1-Q)^{n+1}}{1-Q+Q^2}$.

Answer 2

Well, $\sum_{k=0}^n q^k = \frac{q^{n+1}-1}{q-1}$.

In your case, $q=Q(1-Q)$.