Math induction problem.

Question

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?

Answer 1

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.

Answer 2

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?