Question

Stiind ca:1/1+2+1/1+2+3+............1/1+2+3+........+n=2013/2015 sa se determine n∈N+

Answer 1

$1+ \frac{1}{1+2} + \frac{1}{1+2+3} +...+ \frac{1}{\sum\limits_{k=1}^{n}k} =2( \frac{1}{1\cdot2} + \frac{1}{2\cdot3} +...+ \frac{1}{n(n+1)} )=2(1- \frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}=2\cdot\frac{n}{n+1}$
Continua tu.