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I have been asked to prove that:

$$ \frac{1}{3*4} +\frac{1}{4*5} +\frac{1}{5*6}+...+\frac{1}{(n+2)(n+3)}=\frac{n}{3(n+3)} $$

Here is what I have so far but cannot work out how to go further.

RTP: $$ \frac{1}{3*4} +\frac{1}{4*5} +\frac{1}{5*6}+...+\frac{1}{(n+2)(n+3)}=\frac{n}{3(n+3)} $$

Proof:

When n = 1 $$ \frac{1}{(1+2)(1+3)}= \frac{1}{3(1+3)} $$ $$ \frac{1}{12} = \frac{1}{12} $$

Hence true for n = 1

Assume true for n = k $$ \frac{1}{3*4} +\frac{1}{4*5} +\frac{1}{5*6}+...+\frac{1}{(k+2)(k+3)}=\frac{k}{3(k+3)} $$

When n = (k + 1)

$$ \frac{1}{3*4} +\frac{1}{4*5} +\frac{1}{5*6}+...+\frac{1}{(k+2)(k+3)}+\frac{1}{((k+1)+2)((k+1)+3)}=\frac{k}{3(k+3)}+\frac{1}{((k+1)+2)((k+1)+3)} $$ $$ =\frac{k}{3(k+3)}+\frac{1}{(k+3)(k+4)} $$ $$ =\frac{k(k+4)}{3(k+3)(k+4)}+\frac{3}{3(k+3)(k+4)} $$ $$ =\frac{k(k+4)+3}{3(k+3)(k+4)} $$

Don't know where to go from here / if I have made any mistakes.

Luke Marsden
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I think it's better without induction: $$\sum_{k=1}^n\frac{1}{(k+2)(k+3)}=\sum_{k=1}^n\left(\frac{1}{k+2}-\frac{1}{k+3}\right)=\frac{1}{3}-\frac{1}{n+3}=\frac{n}{3(n+3)}.$$

You need to prove that $$\frac{k}{3(k+3)}+\frac{1}{(k+3)(k+4)}=\frac{k+1}{3(k+4)}$$ or $$k(k+4)+3=(k+1)(k+3),$$ which is true.

Id est, you have no some mistake.

Michael Rozenberg
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