Question

[Nguyễn Thị Minh Khuê]

Cho A= 1/ 1+2+3 + 1/1+2+3+4 + .....+ 1/1+2+3+4+...+19 Cần gấp ai làm xong và đúng nhất mình sẽ Tick cho

Answer 1

[FL.Hermit]

\(A=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{19.20}{2}}\)

=>    \(A=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{19.20}\)

=>   \(\frac{A}{2}=\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

=>   \(\frac{A}{2}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

=>    \(\frac{A}{2}=\frac{1}{3}-\frac{1}{20}\)

=>   \(\frac{A}{2}=\frac{20-3}{20.3}\)

=>   \(\frac{A}{2}=\frac{17}{60}\)

=>   \(A=\frac{17}{30}\)

VẬY    \(A=\frac{17}{30}\)

Answer 2

[Xyz OLM]

Ta có :\(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+19}\)

\(=\frac{1}{3\times4}\times2+\frac{1}{4\times5}\times2+...+\frac{1}{19\times20}\times2\)

\(=2\times\left(\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{19\times20}\right)=2\times\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(=2\times\left(\frac{1}{3}-\frac{1}{20}\right)=2\times\frac{17}{60}=\frac{17}{30}\)