\(A=1+5+5^2+5^3+5^4+...+5^{19}\)

\(5A=5+5^2+5^3+...+5^{19}+5^{20}\)

\(5A-A=\left(5+5^2+5^3+...+5^{20}\right)-\left(1+5+5^2+...+5^{19}\right)\)

\(\Leftrightarrow5A-A=4A=5^{20}-1\Leftrightarrow4A+1=5^{20}-1+1=5^{20}⋮5\)

=> 4A + 1 là hợp số

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}< 5.\frac{1}{25}+10.\frac{1}{30}+10.\frac{1}{40}\)

\(A< \frac{1}{5}+\frac{1}{3}+\frac{1}{4}< \frac{1}{4}+\frac{1}{3}+\frac{1}{4}=\frac{5}{6}\left(đpcm\right)\)

Câu hỏi của Doãn Thị Thanh Thu - Toán lớp 7 - Học toán với OnlineMath tham khảo

Thank you 

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

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

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