Question

[anh]

1/3 cộng 1/6 cộng 1/10 cộng ....... Cộng 1/x(x cộng 1):2 bằng 2009/2011

Answer 1

[Nguyen Van Huong]

Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2009}{2011}\)

Đặt tổng vế trái là A

Ta có : \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\)

\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right)\div2}\right)\)

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

\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\)

\(\frac{1}{2}A=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{x}-\frac{1}{x+1}\right)\)

\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{x+1}\)

\(A=\left(\frac{1}{2}+\frac{1}{x+1}\right):\frac{1}{2}\)

\(A=1+\frac{1}{\left(x+1\right)\div2}\)

\(\Rightarrow1+\frac{1}{\left(x+1\right)\div2}=\frac{2009}{2011}\)

\(\Rightarrow\frac{1}{\left(x+1\right)\div2}=\frac{2009}{2011}-1=\frac{2009}{2011}-\frac{2011}{2011}=-\frac{2}{2011}\)

\(\Rightarrow-\frac{2}{-\left(x+1\right)}=-\frac{2}{2011}\)

\(\Rightarrow-\left(x+1\right)=2011\)

\(\Rightarrow x+1=-2011\) 

\(\Rightarrow x=-2011-1=-2012\)