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- #2

- Feb 7, 2012

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Let $\displaystyle p_k(a) = \sum_{n=1}^k n(1-a)(1-2a)\cdots\bigl(1-(n-1)a\bigr)$. Once you know the formula for $p_k(a)$, you can prove it by induction. The formula isFind the sum of [tex]n[/tex] terms:

[tex]1+2(1-a) +3(1-a)(1-2a)..........k(1-a)(1-2a)...\{1-(k-1)a\}[/tex]

$$p_k(a) = \frac{1-(1-a)(1-2a)\cdots(1-ka)}a.$$

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- #3

- Feb 14, 2012

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Anyway, I will proceed and please forgive and ignore me if, by any chance, I have put my foot into my mouth by asking the following question.

I don't quite understand how one should be able to 'guess' or should one need some prerequisite knowledge in order to be able to do so?Let $\displaystyle p_k(a) = \sum_{n=1}^k n(1-a)(1-2a)\cdots\bigl(1-(n-1)a\bigr)$. Once you know the formula for $p_k(a)$, you can prove it by induction. The formula is

$$p_k(a) = \frac{1-(1-a)(1-2a)\cdots(1-ka)}a.$$

Having said so, I'll show my workout:

k=1:

sum =1

k=2:

sum = 1+2(1-a)=-2a+3

k=3:

$sum = 1+2(1-a)+3(1-a)(1-2a)$

$\;\;\;\;\;\;\;\;=-2a+3+3(1-3a+2a^2)$

$\;\;\;\;\;\;\;\;=6a^2-11a+6$

k=4:

$sum =1+2(1-a)+3(1-a)(1-2a)+4(1-a)(1-2a)(1-3a)$

$\;\;\;\;\;\;\;\;=6a^2-11a+6+4(1-6a+11a^2-6a^3)$

$\;\;\;\;\;\;\;\;=-24a^3+50a^2-35a+10$

k=5:

$sum =1+2(1-a)+3(1-a)(1-2a)+4(1-a)(1-2a)(1-3a)+5(1-a)(1-2a)(1-3a)(1-4a)$

$\;\;\;\;\;\;\;\;=-24a^3+50a^2-35a+10+5(1-10a+35a^2-50a^3+24a^4)$

$\;\;\;\;\;\;\;\;=120a^4-274a^3+225a^2-85a+15$

I noticed that I don't really have to expand the expression for 3(1-a)(1-2a) as in k=3, sum = 1+2(1-a)+

as I can deduce it from the previous result.

Take for example, if I've sum = 1+2(1-a)=-2a+3, then to deduce the value for 3(1-a)(1-2a), I just take 1+ (the terms written in the reverse order of the previous sum but also I add a factor of a each time I go through all the terms and not to forget to change their signs), i.e. $\displaystyle1-3a+2a^2)$

Now, if given $\displaystyle 1+2(1-a)+3(1-a)(1-2a)+4(1-a)(1-2a)(1-3a)+5(1-a)(1-2a)(1-3a)(1-4a)=120a^4-274a^3+225a^2-85a+15$, I can deduce the expression for $\displaystyle 6(1-a)(1-2a)(1-3a)(1-4a)(1-5a)=1-15a+85a^2-225a^3+274a^4-120a^5$, that's it!

To sum up,

Sn=Sn-1+n(1+ (the terms written in the reverse order of the previous sum but also adding another factor of

Does this help in deducing the formula for $\displaystyle p_k(a)$ as you mentioned in your previous post?

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- #4

- Feb 7, 2012

- 2,799

That is not exactly how I came across the formula, but it is as good a way as any. In particular, you started by looking at $p_k(a)$ for small values of $k$ in order to try to find a pattern for it. That is always the best way to approach a problem like this.Hi Opalg, I was wondering if I'm allowed to ask any follow-up question here because obviously I'm not the OP who started it.

Anyway, I will proceed and please forgive and ignore me if, by any chance, I have put my foot into my mouth by asking the following question.

I don't quite understand how one should be able to 'guess' or should one need some prerequisite knowledge in order to be able to do so?

Having said so, I'll show my workout:

k=1:

sum =1

k=2:

sum = 1+2(1-a)=-2a+3

k=3:

$sum = 1+2(1-a)+3(1-a)(1-2a)$

$\;\;\;\;\;\;\;\;=-2a+3+3(1-3a+2a^2)$

$\;\;\;\;\;\;\;\;=6a^2-11a+6$

k=4:

$sum =1+2(1-a)+3(1-a)(1-2a)+4(1-a)(1-2a)(1-3a)$

$\;\;\;\;\;\;\;\;=6a^2-11a+6+4(1-6a+11a^2-6a^3)$

$\;\;\;\;\;\;\;\;=-24a^3+50a^2-35a+10$

k=5:

$sum =1+2(1-a)+3(1-a)(1-2a)+4(1-a)(1-2a)(1-3a)+5(1-a)(1-2a)(1-3a)(1-4a)$

$\;\;\;\;\;\;\;\;=-24a^3+50a^2-35a+10+5(1-10a+35a^2-50a^3+24a^4)$

$\;\;\;\;\;\;\;\;=120a^4-274a^3+225a^2-85a+15$

I noticed that I don't really have to expand the expression for 3(1-a)(1-2a) as in k=3, sum = 1+2(1-a)+3(1-a)(1-2a)

as I can deduce it from the previous result.

Take for example, if I've sum = 1+2(1-a)=-2a+3, then to deduce the value for 3(1-a)(1-2a), I just take 1+ (the terms written in the reverse order of the previous sum but also I add a factor of a each time I go through all the terms and not to forget to change their signs), i.e. $\displaystyle1-3a+2a^2)$

Now, if given $\displaystyle 1+2(1-a)+3(1-a)(1-2a)+4(1-a)(1-2a)(1-3a)+5(1-a)(1-2a)(1-3a)(1-4a)=120a^4-274a^3+225a^2-85a+15$, I can deduce the expression for $\displaystyle 6(1-a)(1-2a)(1-3a)(1-4a)(1-5a)=1-15a+85a^2-225a^3+274a^4-120a^5$, that's it!

To sum up,

Sn=Sn-1+n(1+ (the terms written in the reverse order of the previous sum but also adding another factor ofaeach time we go through all the terms and not to forget to change their signs)

Does this help in deducing the formula for $\displaystyle p_k(a)$ as you mentioned in your previous post?

After staring at $p_2(a)$, $p_3(a)$ and $p_4(a)$ for a while, and not finding any obvious pattern, I happened to notice that (for those small values of $k$) $p_k(1) = 1$, $p_k(1/2) = 2$, $p_k(1/3)=3$ and so on up to $p_k(1/k)=k.$ In other words, $p_k(a) - \frac1a=0$ when $a = 1,\;1/2,\ldots,1/k$. Therefore, by the factor theorem, $p_k(a) - \frac1a$ is a multiple of $(a-1)\bigl(a-\frac12\bigr)\cdots(a-\frac1k\bigr)$. From there, it was quite easy to arrive at the formula for $p_k(a).$

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- #5

- Feb 14, 2012

- 3,931

Awesome! Awesome!That is not exactly how I came across the formula, but it is as good a way as any. In particular, you started by looking at $p_k(a)$ for small values of $k$ in order to try to find a pattern for it. That is always the best way to approach a problem like this.

After staring at $p_2(a)$, $p_3(a)$ and $p_4(a)$ for a while, and not finding any obvious pattern, I happened to notice that (for those small values of $k$) $p_k(1) = 1$, $p_k(1/2) = 2$, $p_k(1/3)=3$ and so on up to $p_k(1/k)=k.$ In other words, $p_k(a) - \frac1a=0$ when $a = 1,\;1/2,\ldots,1/k$. Therefore, by the factor theorem, $p_k(a) - \frac1a$ is a multiple of $(a-1)\bigl(a-\frac12\bigr)\cdots(a-\frac1k\bigr)$. From there, it was quite easy to arrive at the formula for $p_k(a).$

Opalg, I'm not sure if words could accurately describe how grateful I am to you.

I must also admit that I'm tired and worn out for always trying to look for a pattern for questions like this. Hence, what you told me was like an awakening...

THANKS, Opalg!(h)