Question

[Ngô Nam Khánh]

A = \(\frac{1}{1.300}\)+   \(\frac{1}{2.301}\)+    \(\frac{1}{3.302}\)+  . . . . . .+   \(\frac{1}{101.400}\)

B = \(\frac{1}{1.102}\)+    \(\frac{1}{2.103}\) +    \(\frac{1}{3.104}\)+  . . . . . . +   \(\frac{1}{299.400}\)

Tính \(\frac{A}{B}\)

Answer 1

[ST]

Ta có:

A = \(\frac{1}{1.300}+\frac{1}{2.301}+...+\frac{1}{101.400}\)

= \(\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\)

= \(\frac{1}{299}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)

B = \(\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}\)

= \(\frac{1}{101}\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)

= \(\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+...+\frac{1}{400}\right)\right]\)

= \(\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)+\left(\frac{1}{102}+\frac{1}{103}+..+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+..+\frac{1}{299}\right)+\left(\frac{1}{300}+\frac{1}{301}+..+\frac{1}{400}\right)\right]\)

= \(\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{299}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+..+\frac{1}{400}\right)\right]}{\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}=\frac{1}{\frac{299}{\frac{1}{101}}}=\frac{1}{299}\cdot\frac{101}{1}=\frac{101}{299}\)

Answer 2

[TRỊNH ANH TUẤN]

\(\frac{A}{B}=\frac{101}{299}\)