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- Thread starter marobin
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- #1

- Feb 1, 2012

- 57

Well it reduces to $\sum\limits_{i=2}^n (i)x^{n-i}=2x^{n-2}+3x^{n-3}\cdots+(n-1)x+n$I have the following summation and I'm attempting to remove the summation notation. It appears to be the sum of a geometric series but I'm having a great deal of trouble with it. X is an unknown constant.

$$\sum\limits_{i=2}^n (n - (n-i))x^{n-i}$$

But that is not very much help.

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- #4

- Jan 31, 2012

- 54

I have the following summation and I'm attempting to remove the summation notation. It appears to be the sum of a geometric series but I'm having a great deal of trouble with it. X is an unknown constant.

$$\sum\limits_{i=2}^n (n - (n-i))x^{n-i}$$

Thanks.

$$\sum\limits_{i=2}^n (n - (n-i))x^{n-i}=\sum\limits_{i=2}^n ix^{n-i}=x^n\sum\limits_{i=2}^n \frac{i}{x^i}$$

$$\sum\limits_{i=2}^n \frac{i}{x^i}=2(\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}+...+\frac{1}{x^n})$$

$$+(\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}+...+\frac{1}{x^n})+(\frac{1}{x^4}+\frac{1}{x^5}+\frac{1}{x^6}+...+\frac{1}{x^n})+...+\frac{1}{x^n}$$

Could you proceed?

Last edited:

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- #6

$$\sum\limits_{i=2}^n \frac{i}{x^i}=2(\frac{1}{x^2}+\frac{1}{x^3}+\frac{ 1}{x^4}+\frac{1}{x^5}+...+\frac{1}{x^n})+(\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}+...+ \frac{1}{x^n})+(\frac{1}{x^4}+\frac{1}{x^5}+\frac{ 1}{x^6}+...+\frac{1}{x^n})+...+\frac{1}{x^n}$$

in terms of why only the first set of summations is multiplied by 2. Sorry, I'm very rusty on the rules of summation.

- Jan 31, 2012

- 54

$$\sum\limits_{i=2}^n \frac{i}{x^i}=2(\frac{1}{x^2}+\frac{1}{x^3}+\frac{ 1}{x^4}+\frac{1}{x^5}+...+\frac{1}{x^n})+(\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}+...+ \frac{1}{x^n})+(\frac{1}{x^4}+\frac{1}{x^5}+\frac{ 1}{x^6}+...+\frac{1}{x^n})+...+\frac{1}{x^n}$$

in terms of why only the first set of summations is multiplied by 2. Sorry, I'm very rusty on the rules of summation.

I'm little late to work, so just try to answer my question:

What is $$\sum\limits_{i=2}^n \frac{i}{x^i}$$ ?

$$\sum\limits_{i=2}^n \frac{i}{x^i}=\frac{2}{x^2}+\frac{3}{x^3}+\frac{4}{x^4}+...+\frac{n}{x^n}$$

Yes?

So, why then the above equals to what I wrote in my previous post?

Good-luck!

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- #8

- Jan 27, 2012

- 95

Let $\displaystyle S_n=\sum_{i=2}^{n}\frac{i}{x^n}=\frac{2}{x^2}+ \frac{3}{x^3}+\cdots \frac{n}{x^n} \quad (1)$

Multiply both sides by $\frac{1}{x}$:

$\displaystyle \frac{S_n}{x}=\sum_{i=2}^{n}\frac{i}{x^{i+1}}= \frac{2}{x^3}+\frac{3}{x^4}+\cdots \frac{n}{x^{n+1}} \quad (2)$

Subtract (2) from (1):

$$ S_n \left( 1-\frac{1}{x}\right)=\frac{2}{x^2}+ \left( \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}+\cdots \frac{1}{x^n}\right)-\frac{n}{x^{n+1}}$$

Can you proceed?

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- #10

By applying the formula and if the lower bound was 1 instead of 2, I believe its closed form would be: $$\frac{-(n+1)(\frac{1}{x})^{n+1}+n(\frac{1}{x})^{n+2}+x}{(x-1)^{2}}$$

I wanted to make sure I'm taking the correct approach before attempting to see how the formula would change when taking into account that the lower bound is 2 and not 1.