0

How to prove the following with induction?

$$\sum_{k=1}^{2n} \frac{1}{k(k+1)} = \frac{2n}{2n+1}$$

I have difficulty solving this example. I got past base part where I prove that $L(1) = P(1)$ but I cant prove it for $n+1$. Any ideas how to deal with this kind of example?

wythagoras
  • 25,026
Flurrih
  • 11
  • Please refrain from using pictures or links to state your problem. – wythagoras Sep 13 '15 at 09:08
  • Hi and welcome to Math.SE! Have you tried to search this site for similar problems? There are tons of them... Also, this sum is telescoping so someone will probably tell you to solve it using that property instead. – mickep Sep 13 '15 at 09:08
  • Thanks, I tried using to write the formula but i couldnt put the "2n" above sum. mickep thanks for your reply, but its not the same. theres "2n" not "n" which confuse me – Flurrih Sep 13 '15 at 09:11

2 Answers2

1

So here what I got, correct me if I'm wrong...

$$=\frac{2n(2n+2)(2n+3) + 4n +4 }{(2n+1)(2n+2)(2n+3)} = \frac{8n^3 + 20n^2 + 16n +4}{(2n+1)(2n+2)(2n+3)} = \frac{4(n+1)(2n^2+3n+1) }{(2n+1)(2n+2)(2n+3)} = \frac{4(n+1)(n+1)(2n+1)}{(2n+1)(2n+2)(2n+3)} = \frac{(2n+2)(2n+2)(2n+1)}{(2n+1)(2n+2)(2n+3)} = \frac{2n+2}{2n+3}$$ which is equal to P(n+1) so it ends the prove. Did i get it right?

Flurrih
  • 11
0

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

Now try to do a bit of fraction arithmetic.

wythagoras
  • 25,026