Question

Tìm x biết rằng :

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=1\frac{1991}{1993}\)

Answer 1

\(\Leftrightarrow\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+1\right):2}=\dfrac{1991}{1993}\)

\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1991}{1993}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{1991}{1993}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1991}{3986}\)

=>1/x+1=1/1993

=>x+1=1993

hay x=1992