cmr:1+1/2+1/3+...+1/62+1/63+1/64>4
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ND
0
LD
20 tháng 7 2016
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
NH
2
NH
4 tháng 3 2016
các bạn giúp mình nhé, người làm nhanh và đúng sẽ được mình k nhé
7 tháng 4 2024
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
NT
7
25 tháng 3 2016
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
CT
0
19 tháng 4 2019
Đặt \(A=1+\frac{1}{2}+...+\frac{1}{64}\)
Ta có: \(A=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{32}\right)\)\(+\left(\frac{1}{33}+...+\frac{1}{64}\right)\)
Ta thấy : \(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
\(\frac{1}{5}+...+\frac{1}{8}>\frac{1}{8}+...+\frac{1}{8}=\frac{1}{8}.4=\frac{1}{2}\)
\(\frac{1}{9}+\frac{1}{16}>\frac{1}{16}+...+\frac{1}{16}=\frac{1}{16}.8=\frac{1}{2}\)
\(\frac{1}{17}+...+\frac{1}{32}>\frac{1}{32}+...+\frac{1}{32}=\frac{1}{32}.16=\frac{1}{2}\)
\(\frac{1}{33}+...+\frac{1}{64}>\frac{1}{64}+...+\frac{1}{64}=\frac{1}{64}.32=\frac{1}{2}\)
\(\Rightarrow A>1+\frac{1}{2}.6=4\)
Vậy \(A>4\)
A
19 tháng 4 2019
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}+\frac{1}{64}>4.\)
Có :\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}+\frac{1}{64}\)
\(=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+....+\frac{1}{32}\right)+\left(\frac{1}{33}+...+\frac{1}{64}\right)\)
Ta có \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}\)
và \(\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}>\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\)
ta có \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}=\frac{533}{840}\)
và \(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\)
tương tự như trên ta tính được
\(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{16}>\frac{1}{16}\cdot8=\frac{1}{2}\)
\(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}>\frac{1}{32}\cdot16=\frac{1}{2}\)
\(\frac{1}{33}+\frac{1}{34}+\frac{1}{35}+..+\frac{1}{64}>\frac{1}{64}\cdot32=\frac{1}{2}\)
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}+\frac{1}{64}>1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+1\)
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}+\frac{1}{64}>4\)