giải chi tiết giúp bé vs ạ $9^*.Tìm~x\in N^*$ biết: $\frac1{2.5}+\frac1{5.8}+\frac1{8.11}+...+\frac1{x.(x+3)}=A$,Với $A=(\frac1{2^2}-1).(\frac1{3^2}-1).(\frac1{4^2}-1).(\frac1{5^2}-1).(-\frac12)^2$

Question

Accepted Answer

$\displaystyle A=\left(\frac{1}{2^{2}} -1\right) .\left(\frac{1}{3^{2}} -1\right) .\left(\frac{1}{4^{2}} -1\right) .\left(\frac{1}{5^{2}} -1\right) .\left( -\frac{1}{2^{2}}\right)^{2}$
$\displaystyle A=\left(\frac{1}{2} -1\right)\left(\frac{1}{2} +1\right)\left(\frac{1}{3} -1\right)\left(\frac{1}{3} +1\right)\left(\frac{1}{4} -1\right)\left(\frac{1}{4} +1\right)\left(\frac{1}{5} -1\right)\left(\frac{1}{5} +1\right)\left( -\frac{1}{2^{2}}\right)^{2}$
$\displaystyle A=\frac{-1}{2} .\frac{3}{2} .\frac{-2}{3} .\frac{4}{3} .\frac{-3}{4} .\frac{5}{4} .\frac{-4}{5} .\frac{6}{5} .\frac{1}{4}$
$\displaystyle A=\frac{3}{20}$
⟹$\displaystyle \frac{1}{2.5} +\frac{1}{5.8} +\frac{1}{8.11} +...+\frac{1}{x( x+3)} =\frac{3}{20}$
$\displaystyle \Longrightarrow \frac{3}{2.5} +\frac{3}{5.8} +\frac{3}{8.11} +...+\frac{3}{x( x+3)} =\frac{9}{20}$
$\displaystyle \Longrightarrow \frac{1}{2} -\frac{1}{5} +\frac{1}{5} -\frac{1}{8} +\frac{1}{8} -\frac{1}{11} +...+\frac{1}{x} -\frac{1}{x+3} =\frac{9}{20}$
$\displaystyle \Longrightarrow \frac{1}{2} -\frac{1}{x+3} =\frac{9}{20}$
$\displaystyle \Longrightarrow \frac{1}{x+3} =\frac{1}{20}$
$\displaystyle \Longrightarrow x=17$