Ta có:A=1/21+1/22+1/23+...+1/40(có 20 số hạng)

           A>1/40+1/40+...+1/40

           A>20/40=1/2(1)

           A=1/21+1/22+1/23+...+1/40(có 20 số hạng)

           A<1/20+1/20+1/20+...+1/20

           A<20/20=1(2)

Từ (1) và (2)=>1/2<A<1

Ta có :A=1/2+1/2^2+1/2^3+...+1/2^100

         2A=1+1/2+1/2^2+...+1/2^99

         2A-A=(1+1/2+1/2^2+...+1/2^99)-(1/2+1/2^2+1/2^3+...+1/2^100)

          A=1-1/2^100

Dễ thấy A>0 và 1-1/2^100<1

=>0<A<1

Có F = \(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)

\(\Rightarrow F=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}=\frac{15}{14}>1\left(1\right)\)

\(F=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}< 2=\frac{20}{10}\left(2\right)\)

Từ ( 1 ),( 2 ) \(\Rightarrow1< F< 2\)

Vậy \(1< F< 2\left(đpcm\right)\)

P.S: đpcm: Điều phải chứng minh

 \(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\)

F < \(\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}< \frac{20}{10}=2\)

ai nhanh mình k

1 /2 -1 /4 + 1 /8-1 /16 + 1 /32-1 /64 < 1 /3

Cách 1:21/64 < 1/3

Cách 2:21/64 < 0.(3)

Đúng

1 /2 + 1 /4 + 1 /8 + 1 /16 + 1 /32 + 1 /64 < 1 /3

Cách 2:63/64 < 0.(3)

Ko đúng

Câu 3 mình ko biết

a)cho \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)là A

ta có:A=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)

2A=\(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\right)2\)

2A=\(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)

2A+A=\(\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\right)\)

3A=\(1-\frac{1}{64}\Rightarrow3A=\frac{63}{64}\Rightarrow A=\frac{21}{64}< \frac{1}{3}\)

vậy \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)

b) sai đề (\(\frac{63}{64}< \frac{1}{3}\)hay sao)

c)sai nối (nếu x=y=3 thì 2x+3y=17 chia hết nhưng 9x+5y=42 ko chia hết)

giúp mình nhé

\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}+\dfrac{1}{64}\\ =\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{17}+\dfrac{1}{18}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{33}+\dfrac{1}{34}+...+\dfrac{1}{64}\right)\)

Ta thấy:

\(\dfrac{1}{3}\) lớn hơn \(\dfrac{1}{4}\)

\(\dfrac{1}{5};\dfrac{1}{6};\dfrac{1}{7}\) lớn hơn \(\dfrac{1}{8}\)

\(\dfrac{1}{9};\dfrac{1}{10};...;\dfrac{1}{15}\) lớn hơn \(\dfrac{1}{16}\)

\(\dfrac{1}{17};\dfrac{1}{18};...;\dfrac{1}{31}\) lớn hơn \(\dfrac{1}{32}\)

\(\dfrac{1}{33};\dfrac{1}{34};...;\dfrac{1}{63}\) lớn hơn \(\dfrac{1}{64}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>\dfrac{1}{2}+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{32}+\dfrac{1}{32}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{64}+\dfrac{1}{64}+...+\dfrac{1}{64}\right)\\ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\\ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>3\)

Vậy \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>3\)(ĐPCM)

Đặt tổng trên là A

\(2A=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\)

\(3A=2A+A=1-\dfrac{1}{64}< 1\Rightarrow A< \dfrac{1}{3}\)

Ta có: A = 1/2+1/3+1/4+...+1/62+1/63+1/64

A = 1+(1/2+1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+1/10+...+1/16)+...+(1/17+1/18+....+1/32)+(1/33+1/34+...+1/64)

Ta có: 1/2+1/3+1/4>1/2+1/4+1/4=1

1/5+1/6+1/7+1/8>1/8+1/8+1/8+1/8=1/8.4=1/2

1/9 +1/10+...+1/16>1/16+1/16+...1/16=1/16.8=1/2

1/33+1/34+...+1/64>1/64+1/64+...+1/64=1/64.32=1/2

Vậy A > 4

Xin ai giải hộ cái

Ta có A = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/64

          A = 1 + (1/2 + 1/3 + 1/4) + (1/5 + 1/6 + ... + 1/8) + (1/9 + 1/10 + 1/11 + ... + 1/16) + (1/17 + 1/18 + 1/19 + ... + 1/32) + (1/33 + 1/34 + 1/35 + ... + 1/64)

=> A > 1 +  (1/2 + 1/4.2) + 1/8.4 + 1/16.8 + 1/32.16 + 1/64.32

     A > 1 + 1 + 1/2 + 1/2 + 1/2 + 1/2

    A > 4 (DPCM).