Question

[Vy 's 's Hachico]

tìm tỉ số của A và B, biết rằng:

A= \(\dfrac{1}{1\times1981}+\dfrac{1}{2\times1982}\)\(+...+\dfrac{1}{n\left(1980+n\right)}+...+\)\(\dfrac{1}{25\times2005}\)

(A có 25 số hạng)

B= \(\dfrac{1}{1\times26}+\dfrac{1}{2\times27}\) +\(...+\dfrac{1}{m\left(m+25\right)}+...+\)\(\dfrac{1}{1980\times2005}\)

(b có 1980 số hạng)

Answer 1

[Akai Haruma]

Lời giải:

Ta có \(A=\frac{1}{1.1981}+\frac{1}{2.1982}+...+\frac{1}{25.2005}\)

\(\Rightarrow 1980A=\frac{1980}{1.1981}+\frac{1980}{2.1982}+...+\frac{1980}{25.2005}\)

\(\Leftrightarrow 1980A=\frac{1981-1}{1.1981}+\frac{1982-2}{2.1982}+....+\frac{2005-25}{25.2005}\)

\(\Leftrightarrow 1980A=1-\frac{1}{1981}+\frac{1}{2}-\frac{1}{1982}+...+\frac{1}{25}-\frac{1}{2005}\)

\(1980A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)-\left(\frac{1}{1981}+\frac{1}{1982}+..+\frac{1}{2005}\right)\) (1)

Lại có:

\(25B=\frac{25}{1.26}+\frac{25}{2.27}+...+\frac{25}{1980.2005}\)

\(\Leftrightarrow 25B=\frac{26-1}{1.26}+\frac{27-2}{2.27}+...+\frac{2005-1980}{1980.2005}\)

\(\Leftrightarrow 25B=1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+...+\frac{1}{1980}-\frac{1}{2005}\)

\(25B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1980}\right)-\left(\frac{1}{26}+\frac{1}{27}+....+\frac{1}{2005}\right)\)

\(25B=\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{1981}+\frac{1}{1982}+...+\frac{1}{2005}\right)\) (2)

Từ \((1); (2)\Rightarrow 1980A=25B\Rightarrow \frac{A}{B}=\frac{25}{1980}=\frac{5}{396}\)