How can I simplify a summation with constant variables A and B?

[nobahar]

Homework Statement

Hello!
I'm guessing this is precalculus.

There is an intermediate step in a simplifying process and I got to:
\sum_{x=1}^n<br /> xA^{-Bx} Where A is a constant and B is a constant.

Homework Equations

I was wondering how to write this without the summation sign (similar to sum of squares, etc.):
\sum_{x=1}^n<br /> xA^{-Bx}
(Same equation as above).

The Attempt at a Solution

I wolfram alphaed it. But it doesn't show the steps, plus I don't use it very often and I'm not sure if it interpreted it the way I meant; I think it did.
I tried a similar thing to the sum of consecutive integers approach, and I tried using the natural log to get rid of the x in the power, but to no avail. How would I begin to approach this problem? Is this one sufficiently complicated that it should be left to looking up on wolfram alpha?!

Any help appreciated.

 

[Millennial]

I am laughing at myself for not seeing an easy solution and instead resorting to integration.
The easy solution lies on the fact that \displaystyle \sum_{k=1}^{n}kA^{-Bk} is simply \displaystyle \sum_{k=1}^{n}k(A^{-B})^k, and this has a general formula which I will give here:
\sum_{k=1}^{n}kr^k=\frac{r(1-r^{n+1})}{(1-r)^2}-\frac{(n+1)r^{n+1}}{1-r}
What is the answer to your problem then?

 

[nobahar]

Hello Millennial, many thanks for the response.
I wouldn't have been able to solve it that way as I didn't know there was a general solution like that. Do you know where I can find the proof, or what it's 'called'?
Thanks again!

 

[Mentallic]

Hello Millennial, many thanks for the response.
I wouldn't have been able to solve it that way as I didn't know there was a general solution like that. Do you know where I can find the proof, or what it's 'called'?
Thanks again!

To prove that S=\sum_{k=1}^{n}kr^k=\frac{r(1-r^{n+1})}{(1-r)^2}-\frac{(n+1)r^{n+1}}{1-r}

Consider \int S dr = \sum_{k=1}^{n}\frac{k}{k+1}r^{k+1}
= \sum_{k=1}^{n}\left(1-\frac{1}{k+1}\right)r^{k+1}
= r\left(\sum_{k=1}^{n}r^k-\sum_{k=1}^{n}\frac{r^k}{k+1}\right)

Now, you should know the formula for \sum_{k=1}^{n}r^k and transform the second sum such that the denominator is k and then after you take the derivative the denominator should be canceled out.

 

[LCKurtz]

I will add to Mentallic's answer. He shows how to verify the answer after you know what it is. Assuming you don't know the answer ahead of time, you could approach it like this. To simplify the notation I will let $r = A^{-b}$ like he did and I will use $k$ for the index of summation. You have$$
\sum_{k=1}^n kr^k =r\sum_{k=1}^n kr^{k-1} =r\sum_{k=1}^n \frac d {dr} r^k
= r\frac d {dr}\sum_{k=1}^n r^k=r\frac d {dr}\left(\frac{r-r^{n+1}}{1-r}\right)$$Now just differentiate and simplify to get the answer.

 

[PrashntS]

Hello Millennial, many thanks for the response.
I wouldn't have been able to solve it that way as I didn't know there was a general solution like that. Do you know where I can find the proof, or what it's 'called'?
Thanks again!

As far as I know, it is called AGP, arithmetico-geometric progression.

 

[Mentallic]

I will add to Mentallic's answer. He shows how to verify the answer after you know what it is. Assuming you don't know the answer ahead of time, you could approach it like this. To simplify the notation I will let $r = A^{-b}$ like he did and I will use $k$ for the index of summation. You have$$
\sum_{k=1}^n kr^k =r\sum_{k=1}^n kr^{k-1} =r\sum_{k=1}^n \frac d {dr} r^k
= r\frac d {dr}\sum_{k=1}^n r^k=r\frac d {dr}\left(\frac{1-r^{n+1}}{1-r}\right)$$Now just differentiate and simplify to get the answer.

That's actually much simpler.

 

[LCKurtz]

That's actually much simpler.

Except for a typo in the last parentheses, which I have corrected.

 

[nobahar]

= \sum_{k=1}^{n}\left(1-\frac{1}{k+1}\right)r^{k+1} ...and transform the second sum such that the denominator is k and then after you take the derivative the denominator should be canceled out.

Hi Mentallic, many thanks for the response. I couldn't change (k+1) to k without making the sum more complicated. Instead, I didn't factor out the r for the second sum and just took the derivative of \sum_{k=1}^{n}\frac{1}{k+1}r^{n+1}. If I didn't make any mistakes I think it gives the same answer.

He shows how to verify the answer after you know what it is.

Hi Kurtz, thanks for the response to you aswell. I don't understand what you mean here, why is it verification?

$$\sum_{k=1}^n kr^k =r\sum_{k=1}^n kr^{k-1} =r\sum_{k=1}^n \frac d {dr} r^k
= r\frac d {dr}\sum_{k=1}^n r^k=r\frac d {dr}\left(\frac{r-r^{n+1}}{1-r}\right)$$Now just differentiate and simplify to get the answer.

Thankyou Mentallic and Kurtz for these solutions. If its not too much trouble, may I ask how you went about solving the problem? Using derivatives and integrals to simplify these summations is not something that occurred to me. Do you just 'recognise'/'deduce' that this would be a way of finding the solution or is it through experience (I guess it's probably both)?, because there must be many potential approaches to attempting to solve the problem. I ask because I would like to be able to at least make a reasonable attempt at solving them, and I am often at a loss as to how to start thinking about them.
Any help appreciated, thanks for all the help so far!

As far as I know, it is called AGP, arithmetico-geometric progression.

Thanks PrashntS!

 

[LCKurtz]

Hi Kurtz, thanks for the response to you aswell. I don't understand what you mean here, why is it verification?

If I had looked more closely at what he did, I wouldn't have called it verification.

Thankyou Mentallic and Kurtz for these solutions. If its not too much trouble, may I ask how you went about solving the problem?

For me, the $kr^k$ immediately suggests a derivative since had the exponent been $k-1$, it would be the derivative of $r^k$. That was the key idea.

 

[nobahar]

Thanks Kurtz, much appreciated.