Question

Tìm số nguyên dương x để:

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

Answer 1

Lời giải:
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+....+\frac{1}{\frac{x(x+1)}{2}}\)

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

\(=1+2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{(x+1)-x}{x(x+1)}\right)\)

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

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

Ta có: $2-\frac{2}{x+1}=2$

$\Leftrightarrow \frac{2}{x+1}=0$ (vô lý)

Vậy không tồn tại $x$ nguyên dương thỏa mãn.