# Telescoping sums

#### GetReal

##### New member
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

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#### MarkFL

##### Super Moderator
Staff member
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?

Now it is clear