Question

rezolva ecuatia 1+1 /1+2+1/1+2+3+...+1/1+2+3+...+x = 200/100

Answer 1

[tex]\dfrac{1}{1} + \dfrac{1}{1+2} +\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+x} = \dfrac{200}{100} \\ \\ \dfrac{1}{\dfrac{1\cdot 2}{2}}+\dfrac{1}{\dfrac{2\cdot 3}{2}}+\dfrac{1}{\dfrac{3\cdot 4}{2}}+....+\dfrac{1}{\dfrac{x\cdot (x+1)}{2}} = \dfrac{200}{100} \\ \\ \\ \dfrac{2}{1\cdot 2}+\dfrac{2}{2\cdot 3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{x\cdot (x+1)} = \dfrac{200}{100} \Big|:2 \\ \\ \dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x\cdot (x+1)} = \dfrac{100}{100} \\ \\ [/tex]

$\dfrac{2-1}{1\cdot 2}+\dfrac{3-2}{2\cdot 3}+\dfrac{4-3}{3\cdot4}+...+\dfrac{(x+1)-x}{x\cdot (x+1)} = 1 \\ \\ \dfrac{2}{1\cdot 2}-\dfrac{1}{1\cdot2}+\dfrac{3}{2\cdot 3}-\dfrac{2}{2\cdot 3}+...+\dfrac{x+1}{x\cdot(x+1)}-\dfrac{x}{x\cdot (x+1)}=1 \\ \\ 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1} = 1 \\ \\ 1-\dfrac{1}{x+1} = 1 \\ \\ \dfrac{1}{x+1} = 0 \\ \\ \\ \Rightarrow \boxed{x\in \emptyset}$