\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}+\frac{1}{50}\)

\(=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(99+1\right).99}{2}}+\frac{1}{50}\)

\(=\frac{2}{\left(2+1\right).2}+\frac{2}{\left(3+1\right).3}+...+\frac{2}{\left(99+1\right).99}+\frac{1}{50}\)

\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)+\frac{1}{50}\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)

\(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2.\left(\frac{50}{100}-\frac{1}{100}\right)+\frac{1}{50}=2.\frac{49}{100}+\frac{1}{50}\)

\(=\frac{49}{50}+\frac{1}{50}=1\)

Ket qua la 1 con neu muon xem cach giai thi vao chtt

Đặt \(B=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}\)

\(B=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{4950}\)

\(\Rightarrow B=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)

\(\Rightarrow B=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(\Rightarrow B=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow B=2\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(\Rightarrow B=2.\frac{49}{100}\)

\(\Rightarrow B=\frac{49}{50}\)

\(\Rightarrow A=B+\frac{1}{50}=\frac{49}{50}+\frac{1}{50}=\frac{50}{50}=1\)

minh nghi =1 hoac =2 

cu the la bang may

= 1.

Cách giải: Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+99}+\frac{1}{50}\)

\(=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+\frac{1}{\frac{\left(4+1\right).4}{2}}+....+\frac{1}{\frac{\left(99+1\right).99}{2}}+\frac{1}{50}\)

\(=\frac{1}{\frac{3.2}{2}}+\frac{1}{\frac{4.3}{2}}+\frac{1}{\frac{5.4}{2}}+....+\frac{1}{\frac{100.99}{2}}+\frac{1}{50}\)

\(=\frac{2}{3.2}+\frac{2}{4.3}+\frac{2}{5.4}+...+\frac{2}{100.99}+\frac{1}{50}\)

\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)+\frac{1}{50}\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)

\(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2.\frac{49}{100}+\frac{1}{50}=\frac{49}{50}+\frac{1}{50}=1\)

\(A=\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+...+\frac{2}{99}-\frac{2}{100}+\frac{1}{50}\)

\(A=\frac{2}{2}-\frac{2}{100}+\frac{1}{50}=1\)

Bài mình nhớ tick nha. Nếu ko hiểu thì tham khảo "bài toán số 76" nha