Find the sum of the first n terms if a series is both geometric and arithmetic

Question

I have a simple series of the form:

$n+q(n-1)+q^2(n-2)+q^3(n-3)+q^4(n-4)...q^{n-1}$

Where $n$ is in the natural numbers, $q$ is a probability between $[0,1]$. I was wondering if there is some way to present this sum elegantly. I assume that two terms are needed.

I think this question is similar to this thread but not quite the same, I'm under the impression the answer is much more complicated...

Answer 1

Let $$F_n(q)=n+q(n-1)+\cdots +q^{n-1}$$

Note: that's what I believe you intended.

Then $$F_n(q)-qF_n(q)=n-q-q^2-\cdots -q^n$$

Now, $$q+q^2+\cdots q^n=q\times (1+q+\cdots +q^{n-1})$$ and the term in parentheses is a standard Geometric Series. Specifically, $$1+q+\cdots +q^{n-1}=\frac {q^n-1}{q-1}$$

It follows that $$F_n(q)=\frac 1{1-q}\times \left( n-q\times \frac {q^n-1}{q-1}\right)$$

Answer 2

Another (and longer) approach: the series is equal to $n$ times a geometric series minus $q$ times a derived geometric series. One gets

$$n \sum_{k=0}^{n-1} q^k - q \sum_{k=1}^{n-1}k q^{k-1}$$ $$= n \frac{1-q^n}{1-q} - q \frac{d}{dq} \left(\sum_{k=0}^{n-1} q^k \right)$$ $$= n \frac{1-q^n}{1-q} - q \frac{1-nq^{n-1} - (n-1)q^{n}}{1-q}$$ $$=\frac{n}{1-q} + \frac{q(1-q^n)}{(1-q)^2} $$