Question

sa se calculeze suma
S=1/(1+2)+1/(1+2+3)+..........................+1/(1+2+3+.............+2009)

Answer 1

1/3+1/6+1/10+1/15+...+2/2009*2010=
2/2*3+2/3*4+2/4*5+2/5*6+ ...+2/2009*2010=
2( 1/2*3+1/3*4+1/4*5+...+1/2009*2010)=
1(1/2-1/3+1/3-1/4+1/4-1/5+....+1/2009-1/2010)=
2(1/2-1/2010)=
2(1005-1)/2010=
1004/1005

Answer 2

$\it S = \dfrac{1}{1+2}+\dfrac{1}{1+2+3} +\dfrac{1}{1+2+3+4} +\ ...\ +\dfrac{1}{1+2+...+2009}$

Aplicăm formula lui Gauss, pentru numitorul fiecărei fracții.

Exemplificăm pentru ultima fracție:

$\it \dfrac{1}{1+2+3+\ ...\ +2009} = \dfrac{1}{\dfrac{2009\cdot2010}{2}} =\dfrac{2}{2009\cdot2010}$

Prin urmare, vom avea:

[tex]\it S = \dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+\ ...\ +\dfrac{2}{2009\cdot2010} = \\\;\\ \\\;\\ = 2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\ ...\ +\dfrac{1}{2009\cdot2010}\right) [/tex]

Fiecare fracție din paranteză se descompune într-o diferență de două fracții, folosind formula :

$\it \dfrac{1}{n(n+1)} = \dfrac{1}{n} -\dfrac{1}{n+1}$

Suma devine:

[tex]\it S = 2\left(\dfrac{1}{2} - \dfrac{1}{3} +\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5} +...+\dfrac{1}{2009}-\dfrac{1}{2010}\right)= \\\;\\ \\\;\\ = 2\left(\dfrac{1}{2} -\dfrac{1}{2010}\right) = 1 - \dfrac{1}{1005} = \dfrac{1004}{1005}[/tex]