Question

Rezolvaţi ecuaţia:
1+1/1+2 +1/1+2+3 +...+ 1/1+2+3+...+x = 200/101 , x aparţine lui N*
"/" - linie de fracţie

Answer 1

$\displaystyle 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots +\frac{1}{1+2+3+\ldots x}=\frac{200}{101}$.
Suma se scrie restrâns
$\displaystyle\sum_{k=1}^x\displaystyle\frac{1}{1+2+\ldots k}=\sum_{k=1}^x\displaystyle\frac{1}{\frac{k(k+1)}{2}}=\\=\sum_{k=1}^x\frac{2}{k(k+1)}=2\sum_{k=1}^x\left(\frac{1}{k}-\frac{1}{k+1}\right)=\\=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\ldots+\frac{1}{x}-\frac{1}{x+1}\right)=\\=2\left(1-\frac{1}{x+1}\right)=\frac{2x}{x+1}$
Se obține ecuația
$\frac{2x}{x+1}=\frac{200}{101}\Rightarrow 202x=200x+200\Rightarrow x=100$

Answer 2

folosesti suma lui Gauss pentru toate sumele
1+2= 2*3/2
1+2+3=3*4/2.........1+2+3+... x=x(x+1)/2
Stii ca 1 supra un nr. inseamna inversul acelui nr. 
deci, ex. devine 1+2/2*3+2/3*4+2/4*5+.......+2/x(x+1)=200/101
Il dam pe 2 factor comun
1+2(1/2*3+1/3*4+1/4*5+........+ 1/x(x+1)=200/101
Stii ca 1/2*3=1/2-1/3
1/3*4=1/3-1/4 s.a m.d
deci, 1+2(1/2-1/3+1/3-1/4+1/4-1/5+....+1/x-1/(x+1))=200/101
in paranteza se reduc fractiile si ramane
1+2(1/2-1/(x+1))=200/101
rezolvi paranteza
1+2(x+1-2supra 2(x+1)=200/101
1+(x-1)/(x+1)=200/101
x-1 supra x+1=200/101
x-1 supra x+1=99/101
x=100