# Question for preparation for the IMC internationals [closed]

(1/(1+2))+(1/(1+2+3))+...+(1/(1+2+3+...+51))

Guys, I'm having difficulties working this one out. Using a calculator, it would have been easy, however in the internationals we will not be allowed one.

Can you guys please help me, as this question will help me solve many other quesions, of similar type.

Kevin

## closed as off-topic by user21820, YuiTo Cheng, GNUSupporter 8964民主女神 地下教會, Cesareo, José Carlos SantosMay 18 at 18:41

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• Work out $1/(1+2)$, $1/(1+2)+1/(1+2+3)$, $1/(1+2)+1/(1+2+3)+1/(1+2+3+4)$. What do you notice? – Lord Shark the Unknown Apr 11 at 15:24
• Sorry, I haven't understood – kenith Apr 11 at 15:25
• The first key is to know that $1+2+3+\cdots+k=\frac{k(k+1)}{2}$ – Thomas Andrews Apr 11 at 15:31

You want the sum \begin{align} \frac1{1+2}+\frac1{1+2+3}+\dots+\frac1{1+2+\dots+51} &=\sum_{k=2}^{51}\frac1{\frac{k}2(k+1)}\\ &=2\sum_{k=2}^{51}\frac1{k(k+1)}\\ &=2\sum_{k=2}^{51}\left(\frac1{k}-\frac1{k+1}\right)\\ &=2\left(\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\dots+\left(\frac1{51}-\frac1{52}\right)\right)\\ &=2\left(\frac12-\frac1{52}\right)\\ &=1-\frac1{26}\\ &=\frac{25}{26}\\ \end{align} In fact one can generalise this to give $$\frac1{1+2}+\dots+\frac1{1+2+\dots+n}=\frac{n-1}{n+1}$$