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How to reduce this sum? $$\left(\frac1{2^0}-\frac1{2^1}\right)+2\left(\frac1{2^1}-\frac1{2^2}\right)+\cdots+(n-1)\left(\frac1{2^{n-2}}-\frac1{2^{n-1}}\right)+n\left(\frac1{2^{n-1}}-\frac1{2^n}\right) $$

I tried to cancel the terms like a telescoping series, but I couldn't do it. So, does someone have an idea to what to do here?

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  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ Oct 8, 2021 at 8:28
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    $\begingroup$ Is the second term a typo ? I do not think that telescoping works here, but you can use $$\frac{1}{2^{n-1}}-\frac{1}{2^n}=\frac{1}{2^n}$$ for every positive integer $n$ $\endgroup$
    – Peter
    Oct 8, 2021 at 8:31
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    $\begingroup$ The hint allows to reduce the sum to $$\sum_{j=1}^n \frac{j}{2^j}$$ I do not know which techniques you are allowed to use to find an explicite formula for this sum , or maybe this is already the desired result. This is the reason why we need more context to give an answer fitting to your level. $\endgroup$
    – Peter
    Oct 8, 2021 at 8:46
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    $\begingroup$ From the hint suggested by @Peter, it would turn out to be a simple arithmetico-geometric series $\endgroup$
    – DatBoi
    Oct 8, 2021 at 8:51
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    $\begingroup$ @blue thanks mate $\endgroup$ Oct 8, 2021 at 8:58

1 Answer 1

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The general term is $$n\left(\frac{1}{2^{n-1}}-\frac{1}{2^{n}}\right)=n\frac{1}{2^{n-1}}-n\frac{1}{2^{n}}=\left(\frac{n-1}{2^{n-1}}-\frac{n}{2^{n}}\right)+\frac{1}{2^{n-1}}$$

Now can you see the telescoping pattern?. Only the term in the parantheses telescopes. The other is just a simple sum of a geometric progression. You can do that using geometric series formula.

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    $\begingroup$ Excellent, I always used more complicated techniques for this sum , but never have seen this approach (+1) ! Now I understand why the sum is written in a seemingly more complicated way. $\endgroup$
    – Peter
    Oct 8, 2021 at 8:56
  • $\begingroup$ @Mr.GandalfSauron thanks a lot $\endgroup$ Oct 8, 2021 at 8:57
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    $\begingroup$ No problem !!!! $\endgroup$ Oct 8, 2021 at 8:59

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