làm sao để viets phân số

Các bạn xem mình giải có đúng không:

  \(1+\frac{1}{2}+........+\frac{1}{63}+\frac{1}{64}\)

\(=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+...........+\)\(\left(\frac{1}{33}+\frac{1}{34}+...........+\frac{1}{64}\right)\)

\(>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+...........+\left(\frac{1}{64}+.....+\frac{1}{64}\right)\)

=\(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)

=4

Vậy \(1+\frac{1}{2}+..............+\frac{1}{64}>4\)

mình cũng chưa làm đc bài này làm thế nào hả bạn?

1/2+1/3+1/4+….+1/63+1/6t4>3
< => (1/2+1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+1/10+…+1/16)+(1/17+1/18+….+1/31)+(1/32+1/33+…..+1/64)>4
Mà 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=1/2
Tương tự ta có 1/9+1/10+…+1/16>8/16=1/2
1/17+1/18+…+1/31>16/31=1/2
Và 1/32+1/33+…+1/64>32/64=1/2

Mai mình cũng phải nộp bài này mà ko biết làm sao bây giờ?

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).

Với mọi k, n Є N+, n ≥ 2 có 1 / (k + 1) + 1 / (k + 2) + ... + 1 / (k + n) < n / (k + 1) 
=> 
1 = 1 
1 / 2 + 1 / 3 < 2 / 2 = 1 
1 / 4 + 1 / 5 + 1 / 6 + 1 / 7 < 4 / 4 = 1 
1 / 8 + ... + 15 < 8 / 8 = 1 
1 / 16 + ... + 1 / 31 < 16 / 16 = 1 
1 / 32 + ... + 1 / 63 < 32 / 32 = 1 
Cộng vế theo vế có 1 + 1 / 2 + ... + 1 / 63 < 6

1+1/2+1/3+1/4+...+1/63=1+(1/2+1/3)+(1/4+1/5+1/6+1/7)+(1/8+1/9+...+1/15)+(1/16+1/17+..,+1/31)+(1/32+1/33+...+1/63)

                                             <1+(1/2+1/2)+(1/4+1/4+1/4+1/4)+(1/8+1/8+...+1/8)+(1/16+1/16+...+1/16)+(1/32+1/32+...+1/32)

                                              <1+1+1+1+1+1=6

\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)

Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)

\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)

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

Vì \(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}

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)