- #1

marksyncm

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## Homework Statement

Prove by induction that ##\sum\limits_{k=1}^{2n} \frac{1}{k(k+1)} = \frac{2n}{2n+1}##

**2. The attempt at a solution**

First I showed that it is true for ##n=1## and ##n=2##. Then, assuming it is true for all ##n##, I attempt to show that it is true for ##n+1##:

$$\sum\limits_{k=1}^{2(n+1)} \frac{1}{k(k+1)} = \frac{2(n+1)}{2(n+1)+1} = ... $$

My problem is what the right side (##...##) of the equation should be. I can't just say that it is equal to ##\frac{2n}{2n+1} + \frac{1}{(n+1)(n+2)}##, because the summation notation requires me to double my ##n## - I understand this to mean that each time we add 1 to ##n##, we are adding

*two additional*terms to the summation total - one is ##2(n+1)-1## and the other is ##2(n+1)##. Based on this reasoning, I concluded that the above equation should be:

$$\sum\limits_{k=1}^{2(n+1)} \frac{1}{k(k+1)} = \frac{2(n+1)}{2(n+1)+1} = \frac{2n}{2n+1} + \frac{1}{(2n+1)(2n+2)} + \frac{1}{(2n+2)(2n+3)}$$

However, I was unable to reduce the right side to resemble the left side, and online algebra tools were of no help. Am I going about this wrong?

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