How i can find the sum of the series? [closed]

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

closed as not a real question by t.b., Did, Grumpy Parsnip, Benjamin Lim, Asaf KaragilaMar 29 '12 at 18:59

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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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