- #1
juantheron
- 247
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Find 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]
[tex]1+2(1-a) +3(1-a)(1-2a)...k(1-a)(1-2a)...\{1-(k-1)a\}[/tex]
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 isjacks said:Find 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]
Opalg said: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
[sp]$$p_k(a) = \frac{1-(1-a)(1-2a)\cdots(1-ka)}a.$$[/sp]
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.anemone said: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.(Worried)
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 of a each 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?
Opalg said: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 $p_k(a).$
Summation, also known as addition, is a mathematical operation that involves adding multiple numbers together to get a total sum.
Summation is typically written using the Greek letter sigma (Σ) followed by the numbers or variables to be added and a lower and upper bound, such as Σn=1^5 n, which means adding numbers from 1 to 5.
Summation is used in many mathematical applications, including calculating areas under curves, finding the average of a set of numbers, and solving series and sequences.
Summation and integration are both mathematical operations that involve adding numbers. However, summation is used for discrete values, while integration is used for continuous values.
Summation can be used in real-world scenarios such as calculating the total cost of items in a shopping cart, determining the average temperature for a week, or finding the total distance traveled on a road trip.