Chứng tỏ rằng 1+1/2+1/3+1/4+...+1/62+1/63+1/64>4
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Những câu hỏi liên quan
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
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
13 tháng 4 2019
A=1+(1/2 + 1/3 + 1/4)+(1/5 + 1/6 + 1/7 + 1/8)+(1/9+...+1/16)+(1/17+...+1/32)+(1/33+...+1/64)
A>1+(1/2 + 1/4 + 1/4)+(1/8+ 1/8+ 1/8+ 1/8)+(1/16+1/16+...+1/16)+(1/64+...+1/64)
A>1 + 1 + 1/2 + 1/2 + 1/2+ 1/2
A>4
HH
28 tháng 7 2018
\(=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{32}\right)+\left(\frac{1}{33}+...+\frac{1}{64}\right)\)
\(=1+\frac{1}{2}+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32\)
\(=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=1+\frac{1}{2}.6\)
\(=1+3\)
\(=4\)
~~ Bố thí cái li.ke ~~
MV
7 tháng 5 2017
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}+\dfrac{1}{64}\\ =\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{17}+\dfrac{1}{18}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{33}+\dfrac{1}{34}+...+\dfrac{1}{64}\right)\)
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)
NH
0
NH
0
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
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
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