\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{50}\)

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\frac{1}{1}-\frac{1}{50}\)

\(A=\frac{49}{50}\)

Vì \(\frac{245}{420}< \frac{245}{294}< \frac{245}{250}\)

Vậy \(\frac{7}{12}< \frac{49}{50}< \frac{5}{6}\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{48.49}\)

\(A< 1-\frac{1}{49}=\frac{48}{49}< \frac{48}{48}< \frac{40}{48}=\frac{5}{6}\)

Hình như phân số cuối sai đề bn nhỉ?

nhung ma ..+1/99.100

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{100}\)

Ta có A =1/1.2+1/3.4+1/5.6+...+1/99.100

=(1/1.2+1/3.4)+(1/5.6+...+1/99.100)

=7/12+(1/5.6+...+1/99.100)>7/12(1)

A=1-1/2+1/3-1/4+1/5-1/6+...+1/99-1/100

=(1+1/3+1/5+...+1/99)-(1/2+1/4+..+1/100)

=(1+1/2+1/3+1/4+..+1/99+1/100)-2(1/2+1/4+....+1/100)    ( Cộng thêm cả 2 vế với 1/2+1/4+..+1/100)

=(1+1/2+1/3+..+1/100)-(1+1/2+..+1/50)

=1/51+1/52+..+1/100

Dãy số trên có 50 số hang 50 chia hết cho 10 nên ta nhóm 10 số vào 1 nhóm

A=(1/51+1/52+..+1/60)+(1/61+1/62+..+1/70)+(1/71+1/72+..+1/80)+(1/81+..+1/90)+(1/91+..+1/100)

<1/50.10+1/60.10+1/70.10+1/80.10+1/90.10=1/5+1/6+1/7+1/8+1/9<1/5+1/6+1/7.3=167/210<175/210=5/6

=>A<5/6(2)

từ 1 và 2 => đpcm