\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}+\frac{1}{50}\)
\(=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(99+1\right).99}{2}}+\frac{1}{50}\)
\(=\frac{2}{\left(2+1\right).2}+\frac{2}{\left(3+1\right).3}+...+\frac{2}{\left(99+1\right).99}+\frac{1}{50}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2.\left(\frac{50}{100}-\frac{1}{100}\right)+\frac{1}{50}=2.\frac{49}{100}+\frac{1}{50}\)
\(=\frac{49}{50}+\frac{1}{50}=1\)
Ket qua la 1 con neu muon xem cach giai thi vao chtt
Đặt \(B=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}\)
\(B=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{4950}\)
\(\Rightarrow B=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)
\(\Rightarrow B=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow B=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow B=2\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\Rightarrow B=2.\frac{49}{100}\)
\(\Rightarrow B=\frac{49}{50}\)
\(\Rightarrow A=B+\frac{1}{50}=\frac{49}{50}+\frac{1}{50}=\frac{50}{50}=1\)
1/1+2=2/2.3
1/1+2+3=2/3.4
....1/1+2+3+...+99=2/99.100
2/2.3+2/3.4+....+2/99.100+1/50
2(1/2.3+1/3.4+...+1/99.100)+1/50
2(1/2-1/3+1/2+1/4+...+1/99-1/100)+1/50
2(1/2-1/100)+1/50
2.49/100+1/50
49/50+1/50
=> 1
Nho tick nha ban cau nay o violympic
