\(\frac{50}{3227}\)

kết quả là:50/3227

Ta có:\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

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

@Ác Mộng ở đoạn cuối tự nhiên bỏ mất số 2 luôn, giải sai rồi kìa

1-2+3-4+5-6+...+51-52+53

=(1-2)+(3-4)+...+(51-52)+53

=(-1)+(-1)+...+(-1)+53

=(-1)×26+53

=-26+53

=27

1-2+3-4+5-6+...+51-52+53

=(1-2)+(3-4)+...+(51-52)+53

=(-1)+(-1)+...+(-1)+53

=(-1)×26+53

=-26+53

=27

Sửa đề: Chứng minh 1/2<1/51+1/52+...+1/100<1

Ta có: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};\ldots;\frac{1}{100}<\frac{1}{50}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}\)

=>\(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}<\frac{50}{50}\)

=>\(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}<1\) (1)

Ta có: \(\frac{1}{51}>\frac{1}{100};\frac{1}{52}>\frac{1}{100};\ldots;\frac{1}{100}=\frac{1}{100}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}\)

=>\(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}>\frac{50}{100}\)

=>\(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}>\frac12\) (2)

Từ (1),(2) suy ra 1/2<\(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\) <1