a)

\(\left(1+\frac{1}{51}\right).\left(1+\frac{1}{52}\right).\left(1+\frac{1}{53}\right)...\left(1+\frac{1}{1963}\right)\)

\(=\frac{52}{51}.\frac{53}{52}.\frac{54}{53}...\frac{1964}{1963}\)

\(=\frac{1964}{51}\)

b)

\(\left(1+\frac{1}{400}\right).\left(1+\frac{1}{401}\right).\left(1+\frac{1}{402}\right)...\left(1+\frac{1}{2010}\right)\)

\(=\frac{401}{400}.\frac{402}{401}.\frac{403}{402}...\frac{2011}{2010}\)

\(=\frac{2011}{400}\)

\(\frac12+\frac{1}{12}+\frac{1}{30}+\cdots+\frac{1}{9900}\)

\(=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{99\cdot100}\)

\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac12+\cdots+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac12+\frac14+\cdots+\frac{1}{100}\right)\)
\(=1+\frac12+\cdots+\frac{1}{99}+\frac{1}{100}-1-\frac12-\cdots-\frac{1}{50}\)

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

Ta có: \(A=\left(\frac12+\frac{1}{12}+\frac{1}{30}+\cdots+\frac{1}{9900}\right):\left(\frac{-6}{51}-\frac{6}{52}-\cdots-\frac{6}{100}\right)\)

\(=\frac{\frac{1}{51}+\frac{1}{52}+\cdots+100}{-6\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\right)}=\frac{-1}{6}\)

Từ 50 đến 99 có 50 số; ta cho tất cả các phân số đó về 1/100; ta có 50/100 = 1/2; còn dư một số phần chênh giữa 1/100 va các phân số đó.

Sửa đề: \(S=\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\)

Ta có: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};\ldots;\frac{1}{75}<\frac{1}{50}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{25}{50}=\frac12\) (1)

Ta có: \(\frac{1}{76}<\frac{1}{75};\frac{1}{77}<\frac{1}{75};\ldots;\frac{1}{100}<\frac{1}{75}\)

Do đó: \(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}<\frac{1}{75}+\frac{1}{75}+\cdots+\frac{1}{75}=\frac{25}{75}=\frac13\) (2)

Từ (1),(2) suy ra \(\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}\right)<\frac12+\frac13\)

=>\(S<\frac56\)

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

to giup cau nhe 

Vi tat ca cac phan so tren deu nho hon 1/2 ne tong do se nho hon 1/2

Neu cau cho la dung hay chon cau tra loi cua minh nhe

Ta thầy từ: 1/51 + 1/52 + 1/53 + 1/54 + .....+ 1/98 + 1/99 mỗi số hạng đều lớn hơn 1/100 Mà tổng trên có (100-51)+1= 50 (số hạng)

Nên 1/51 + 1/52 + 1/53 + 1/54 + .....+ 1/98 + 1/99 + 1/100 > 1/100 x 50 = 50/100 = 1/2 Vậy: s > 1/2