How do I describe the progression $c+ba^{-(n-1)}$? I'm seeking an efficient sum to arbitrary $n$.

Question

How do I characterise a progression of the form: $$ c+ba^{-(n-1)} $$

where integer $n > 0$.

My motivation is that I'm trying to find an efficient sum for this progression, to an arbitrary $n$. (The existing function sums by iterative addition, which is very inefficient for even moderately large $n$.)

From asking earlier I am informed this is not a geometric progression (and hence I can't use the sum formula for a geometric progression).

So, what kind of progression is this, and how do I efficiently sum it, ideally with arithmetic no more complex than exponentiation?

@dmtri, thanks you were faster than my edit :-) The question now uses $n$ in the expression. — bignose, Jan 05 '19 at 06:20
Note that $$\displaystyle\sum_1^n\Big[c+\frac b{a^{k-1}}\Big]=\sum_1^n c+\sum_1^n\frac b{a^{k-1}}=nc+b\Big[\frac{1-a^{-n}}{1-a^{-1}}\Big]$$ where the second term is the sum of a geometric progression. — Shubham Johri, Jan 05 '19 at 06:25
@ShubhamJohri, would you like to put that into an answer? If not, I can do it. — bignose, Jan 06 '19 at 01:22

score 3 · Answer 1 · answered Jan 05 '19 at 06:26

You have $$u_n = c + \frac{b}{a^{n-1}}$$ and you want to find $$S_n = \sum_{k=1}^{n}u_k = \sum_{k=1}^{n}\left( c + \frac{b}{a^{k-1}}\right) =cn+b\left(1+\frac{1}{a^2}+\cdots +\frac{1}{a^{n-1}}\right) $$ $$ = cn+\frac{b}{a^{n-1}}\left(\frac{a^n-1}{a-1}\right).$$

score 0 · Answer 2 · edited Jan 12 '19 at 04:14

0

Note that $$ \sum_{k=1}^n \Big[c+\frac b{a^{k-1}}\Big] = \sum_{k=1}^n c+\sum_{k=1}^n\frac b{a^{k-1}} = nc+b\Big[\frac{1-a^{-n}}{1-a^{-1}}\Big] $$ where the second term is the sum of a geometric progression.

edited Jan 12 '19 at 04:14

bignose

137

answered Jan 06 '19 at 06:21

Shubham Johri

17,659

How do I describe the progression $c+ba^{-(n-1)}$? I'm seeking an efficient sum to arbitrary $n$.

2 Answers2