Formula for finite power series

Question

Are there any formula for result of following power series? $$0\leq q\leq 1$$ $$ \sum_{n=a}^b q^n $$

Answer 1

Yes, consider the following manipulation: $$ S = \sum_{n=a}^b q^n = q^a + q^{a+1} + \cdots + q^b = q^a + q(q^a + a^{a+1} + \cdots + q^{b-1}) = q^a + q(S-q^b). $$ Solving this we find that .... $$ S = \frac{q^a-q^{b+1}}{1-q} $$

Answer 2

Yes. $$ \sum_{n = a}^{b} q^{n} = \sum_{n = 0}^{b} q^{n} - \sum_{n = 0}^{a-1} q^{n}. $$ Now, if $M$ is either $a-1$ or $b$, and if $q<1$ then $$ \sum_{n=0}^{M} q^{n} = {1 - q^{M+1} \over 1 - q} $$ The last formula is explained here: https://en.wikipedia.org/wiki/Geometric_series#Formula

And if $q=1$, then $q^{n} = 1$ for all $n$.

Answer 3

First we assume $a \le b$ (otherwise we swap). With $$ S = \sum_{k=a}^b q^k $$ If $q=1$ we have $S=b - a + 1$. For $q < 1$ we have $$ q S = \sum_{k=a}^b q^{k+1} = \sum_{k=a+1}^{b+1} q^k = S + q^{b+1} - q^a \iff \\ q^a - q^{b+1} = (1-q) S $$ This gives $$ S = \begin{cases} b - a + 1 & \text{ for } q = 1 \\ (q^a-q^{b+1})/(1-q) & \text{ for } q < 1 \end{cases} $$

Answer 4

I know people have already answered this, but laying out each step one-by-one helped me, I hope it helps y'all too,

Take,

$$ S = \sum_{k=a}^b p^k $$

Expanding it,

$$ S = p^a + p^{a+1} + ... + p^{b} $$

Taking $p$ out,

$$ S = p^a + p (p^{a} + p^{a+1} + ... + p^{b-1}) $$

Now add & subtract $p^{b}$,

$$ S =p^a + p (p^{a} + p^{a+1} + ... + p^{b-1} + p^b - p^b) $$

We recognize $S$ & substitute to get,

$$ S =p^a + p (S - p^b) $$

Distribute,

$$ S=p^a + pS - p^{b+1} $$

Rearrange,

$$ S(1 - p) =p^a - p^{b+1} $$

Finally,

$$ S =\frac{p^a - p^{b+1}}{1 - p} $$