\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+49+50}\)
\(=\dfrac{1}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{1}{\dfrac{3\left(3+1\right)}{2}}+\dfrac{1}{\dfrac{4\left(4+1\right)}{2}}+...+\dfrac{1}{\dfrac{50\left(50+1\right)}{2}}\)
\(=\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{50.51}\right).2\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{50}-\dfrac{1}{51}\right).2\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{51}\right).2\)
\(=\dfrac{49}{102}.2\)
\(=\dfrac{49}{51}\)
Lời giải:
Sử dụng công thức:
\(1+2+....+n=\frac{n(n+1)}{2}\)
\(\Rightarrow \frac{1}{1+2+3+...+n}=\frac{2}{n(n+1)}\)
Do đó:
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+49+50}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{50.51}\)
\(\Rightarrow \frac{A}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\)
\(\frac{A}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{51-50}{50.51}\)
\(\frac{A}{2}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}=\frac{1}{2}-\frac{1}{51}\)
\(\Rightarrow A=1-\frac{2}{51}=\frac{49}{51}\)
