Question

Tìm số tự nhiên x biết rằng:

1/2+1/6+1/12+1/20+...+1/x(x+1)=2008/2009

Answer 1

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)

\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)

\(1-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\frac{1}{x+1}=1-\frac{2008}{2009}=\frac{1}{2009}\)

\(\Rightarrow x+1=2009\)

\(\Leftrightarrow x=2008\)