Question

VA ROG URGENT PLS! 26 PUNCTE!

Answer 1

1/1*2/2+1/2*3/2+...+1/n(n+1)/2=4024/2013=>2/1*2+2/2*3+...2/n(n+1)=>2/1-2/2+2/2-2/3+...+2/n-2/n+1=>2/1-2/n+1=>2n/n+1=4024/2013=>n=2012

Answer 2

$\it 1+\dfrac{1}{1+2} +\dfrac{1}{1+2+3}+ ... +\dfrac{1}{1+2+3+ ...+n} =\dfrac{4024}{2013}$

$\it \dfrac{1}{1+2} +\dfrac{1}{1+2+3}+ ... +\dfrac{1}{1+2+3+ ...+n} =\dfrac{4024}{2013} -1$

$\it \dfrac{1}{1+2} +\dfrac{1}{1+2+3}+ ... +\dfrac{1}{1+2+3+ ...+n} =\dfrac{2011}{2013}$

$\it \dfrac{1}{1+2+3+ ... +n} = \dfrac{1}{\dfrac{n(n+1)}{2}} = \dfrac{2}{n(n+1)} \ \ \ (*)$

Aplicăm relația (*) și ecuația devine:

$\it \dfrac{2}{2\cdot3} +\dfrac{2}{3\cdot4}+ ... +\dfrac{2}{n(n+1)} = \dfrac{2011}{2013} \Leftrightarrow$


$\it \Leftrightarrow 2\left(\dfrac{1}{2\cdot3} +\dfrac{1}{3\cdot4}+ ... +\dfrac{1}{n(n+1)}\right) = \dfrac{2011}{2013}\Leftrightarrow$

$\it 2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4} + ...+\dfrac{1}{n}-\dfrac{1}{n+1}\right) =\dfrac{2011}{2013} \Leftrightarrow$

$\it 2(\dfrac{1}{2}-\dfrac{1}{n+1}) = \dfrac{2011}{2013}\Leftrightarrow 2\cdot\dfrac{n+1-2}{2(n+1)}=\dfrac{2011}{2013}\Leftrightarrow \dfrac{n-1}{n+1} =\dfrac{2011}{2013}\Leftrightarrow$

$\it \Leftrightarrow \dfrac{n-1}{n+1} = \dfrac{2012-1}{2012+1} \Leftrightarrow n = 2012$