Question

[Nguyễn Văn Toại]

Cho A = 1/1x101 + 1/2x102 + 1/3x103 + ... + 1/25x125
B = 1/1x26 + 1/2x27 + 1/3x28 + ... + 1/100x125 
tìm thương A/B

Answer 1

[ST]

Ta có: 

A=\(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{25.125}\)

=\(\frac{1}{100}\left(\frac{100}{1.101}+\frac{100}{2.102}+...+\frac{100}{25.125}\right)\)

=\(\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{25}-\frac{1}{125}\right)\)

=\(\frac{1}{100}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]\)

B=\(\frac{1}{1.26}+\frac{1}{2.27}+...+\frac{1}{100.125}\)

=\(\frac{1}{25}\left(\frac{25}{1.26}+\frac{25}{2.27}+...+\frac{25}{100.125}\right)\)

=\(\frac{1}{25}\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+...+\frac{1}{100}-\frac{1}{125}\right)\)

=\(\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{125}\right)\right]\)

=\(\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}\right)-\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]\)

= \(\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]\)

=> \(\frac{A}{B}\)=\(\frac{\frac{1}{100}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]}{\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]}\)=\(\frac{1}{\frac{100}{\frac{1}{25}}}\)=\(\frac{1}{100}\cdot25=\frac{25}{100}=\frac{1}{4}\)