Question

Tìm số nguyên dương x để: \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=2\).

Answer 1

\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=2\)

=> \(1+\frac{1}{\frac{2\left(1+2\right)}{2}}+\frac{1}{\frac{3\left(1+3\right)}{2}}+....+\frac{1}{\frac{x\left(x+1\right)}{2}}=2\)

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

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

=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}\)

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

=> \(\frac{1}{x+1}=0\Rightarrow x\in\varnothing\)

Answer 2

            Bài làm :

Ta có :

\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=2\)

 \(\Leftrightarrow1+\frac{1}{\frac{2\left(1+2\right)}{2}}+\frac{1}{\frac{3\left(1+3\right)}{2}}+....+\frac{1}{\frac{x\left(x+1\right)}{2}}=2\)

 \(\Leftrightarrow1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=2\)

 \(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=1\)

 \(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}\)

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

 \(\Leftrightarrow\frac{1}{x+1}=0\)

=> Không tồn tại x