Question

1)Arătați că 1/10+1/11+1/12+...+1/19 mai mic decât 1
2)Calculați valoarea sumei:
S=(1/1-2)+(1/1-3)+(1/1-4)+...+(1/1-2011)-(1/1+1/2+2/3+3/4+...+2010/2011)
astea sunt tot cu fracții
vă rog să scrieți cât mai repede este urgent!!!!!

Answer 1

1/10+1/11+...+1/19<1/10+1/10+..+1/10=10*1/10=1
a doua, vezi atasament
credca e o mica gresealde tipar in carte
dacv  prima paranteza arfi mers pana la 1/(1-2012)=-1/2011 atunci si fractia ramasa singura, -2010/2011  insumata algebric  cu aceasta ne-afi dat inca un -1 sirezultatul final ar fi fost -2012

asa este -2011-2010/2011=(-2011²-2010)/2011=-4046131/2011

Answer 2

1)
$\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}<\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{10}{10}=1$

2)
$S=\frac{1}{-1}+\frac{1}{-2}+...+\frac{1}{-2010}-\frac{1}{1}-\frac{1}{2}-...-\frac{2010}{2011}$
$=\frac{1}{-1}+\frac{1}{-2}+...+\frac{1}{-2010}-\frac{-1}{-1}-\frac{-1}{-2}-...-\frac{-2010}{-2011}$
$=\frac{1}{-1}-\frac{-1}{-1}+\frac{1}{-2}-\frac{-1}{-2}+\frac{1}{-3}-\frac{-2}{-3}+...+\frac{1}{-2010}-\frac{-2009}{-2010}-\frac{-2010}{-2011}$
$=\frac{2}{-1}+\frac{2}{-2}+\frac{3}{-3}+...+\frac{2010}{-2010}-\frac{-2010}{-2011}$
$=(-2)+(-1)+(-1)+...+(-1)-\frac{-2010}{-2011}$
$=-2011-\frac{-2010}{-2011}$
$=-2011-\frac{2010}{2011}$
$=\frac{2011(-2011)}{2011}-\frac{2010}{2011}$
$=\frac{2011(-2011)-2010}{2011}$
$=\frac{(0+2011)(0-2011)-2010}{2011}$
$=\frac{0^2-2011^2-2010}{2011}$
$=\frac{-(2011^2)-2010}{2011}$
$=\frac{-4044121-2010}{2011}$
$=\frac{-4046131}{2011}$
$=-\frac{4046131}{2011}$