Telescoping sums

GetReal

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Mar 24, 2019
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Part a you just multiply by 7/7 to change the denominator so that you can subtract.
I'm not so sure about the second part

Screen Shot 2019-03-27 at 1.13.23 am.png
 
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MarkFL

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Okay, having proved the given identity, then we know:

\(\displaystyle \frac{k}{7^k}=\frac{7}{36}\left(\frac{6k+1}{7^{k}}-\frac{6(k+1)+1}{7^{k+1}}\right)\)

And so, the sum \(S\) we are asked to simplify becomes:

\(\displaystyle S=\frac{7}{36}\left(\sum_{k=0}^{n-1}\left(\frac{6k+1}{7^{k}}\right)-\sum_{k=0}^{n-1}\left(\frac{6(k+1)+1}{7^{k+1}}\right)\right)\)

Let's re-index the second sum as follows:

\(\displaystyle S=\frac{7}{36}\left(\sum_{k=0}^{n-1}\left(\frac{6k+1}{7^{k}}\right)-\sum_{k=1}^{n}\left(\frac{6k+1}{7^{k}}\right)\right)\)

Now, strip off the first term from the first sum and the last term from the second sum:

\(\displaystyle S=\frac{7}{36}\left(1+\sum_{k=1}^{n-1}\left(\frac{6k+1}{7^{k}}\right)-\sum_{k=1}^{n-1}\left(\frac{6k+1}{7^{k}}\right)-\frac{6n+1}{7^{n}}\right)\)

Can you proceed?
 

Youtra12

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Mar 26, 2019
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Now it is clear
 
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