Question

(x+1)/(1×3)+(x+1)/(3×5)+(x+1)/(5×7)+.....+(x+1)/(2019×2021)=1010/2021​

Answer 1

Răspuns:

$\frac{x+1}{1*3} +\frac{x+1}{3*5} +\frac{x+1}{5*7} +...+\frac{x+1}{2019*2021} =\frac{1010}{2021}$

$(x+1)(\frac{1}{1*3} +\frac{1}{3*5} +\frac{1}{5*7} +...+\frac{1}{2019*2021} )=\frac{1010}{2021}$  /(x+1)

$\frac{1}{1*3} +\frac{1}{3*5} +\frac{1}{5*7} +...+\frac{1}{2019*2021}=\frac{1010}{2021}*\frac{1}{x+1}$ /*2

$\frac{2}{1*3} +\frac{2}{3*5} +\frac{2}{5*7} +...+\frac{2}{2019*2021}=\frac{2020}{2021}*\frac{1}{x+1}$

$\frac{2}{n(n+2)} =\frac{1}{n}-\frac{1}{n+2}$

$\frac{1}{1}- \frac{1}{3}+ \frac{1}{3} -\frac{1}{5}+\frac{1}{5}- \frac{1}{7}+...+\frac{1}{2019}- \frac{1}{2021} =\frac{2020}{2021} *\frac{1}{x+1}$

$\frac{2021)1}{1}-\frac{1}{2021} =\frac{2020}{2021} *\frac{1}{x+1}$

$\frac{2021-1}{2021} =\frac{2020}{2021} *\frac{1}{x+1}$

$\frac{2020}{2021} =\frac{2020}{2021} *\frac{1}{x+1}$  ⇒  $\frac{1}{x+1}=1$ ⇒ x+1=1 ⇒ x=0

Explicație pas cu pas: