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I've got a sequence defined as such:

Between $0$ and $5$, $i = 1$

Between $6$ and $10$, $i = d$

Between $11$ and $15$, $i = d^2$

Between $16$ and $20$, $i = d^3$ ... I guess I could write it like this for $n=5$:

Between $k(n)+1$ and $(k+1)n$, $i= d^k$ and I'm looking for the $n$th term of the induced serie. (ie the sum of the $n$th first terms of my sequence)

Actually, I'm trying to solve a real-world problem here. We're selling products with a progressive discount (eg the first $5$ items are at full price, then the $5$ following items have a $10%$ discount, then there's an extra 10 percent discount for the next clip of $5$, etc).

So far I'm handling this kind of manually.

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I edited out some nonsense, but it's going to take some more work to get this question into decent shape. –  Gerry Myerson Mar 29 '12 at 11:29
'cryptography' tag should be added. –  Salech Alhasov Mar 29 '12 at 11:37
can not understand what he means. –  noname1014 Mar 29 '12 at 12:00
@SalechAlhasov Cryptography? –  user21436 Mar 29 '12 at 12:33
@Kannappan Sampath: Bad sense of humor I guess... :) –  Salech Alhasov Mar 29 '12 at 12:40
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closed as not a real question by t.b., Did, Grumpy Parsnip, Benjamin Lim, Asaf Karagila Mar 29 '12 at 18:59

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2 Answers

The case $d=0$ is not in line with the others. For $d\geqslant1$, the value is $\displaystyle \color{red}{i=d^{\lceil\frac{d}5\rceil-1}}.$

The sum from $n=0$ to $n=5k+\ell$ with $0\leqslant\ell\leqslant4$ is $$ \frac{6-d}{1-d}+d^k\cdot\left(\ell-\frac5{1-d}\right). $$

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So the series is $1+1+1+1+1+d+d+d+d+d+d^2+d^2+d^2+d^2+d^2+d^3...$? I don't think that it can be simplified much when stopping after the n'th term. The only thing you can do is to accumulate every 5 terms block to get $5+5d+5d^2+5d^3+...$.

It might be a questionable business strategy though. Imagine we are puying $5n$ items. These can be accumulated in the form $$5+5d+5d^2+5d^3+...+5d^n = 5(1+d+d^2+d^3+...+d^n) = \frac{5}{1-d}\quad\text{as }n\rightarrow\infty$$

So no matter how many items I buy that have this kind of discount, I will never pay more than $5/(1-d)$. Eg for $d=95\%$ this would be 100 times the price for one item. The company will probably have payed much more for hundret of thounds of those items though.

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