Tính tổng sau
a) Cho \(H=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}\)
Hãy chứng tỏ H>2
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Chứng minh rằng H>2
\(H=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+.......+\frac{1}{63}\)
Ta có: H=(1/2+1/3+1/4)+(1/5+...+1/8)+(1/9+1/16)+(1/17+...+1/63)
=> H=13/12 + (1/5+...+1/8)+(1/9+...+1/16)+(1/17+...+1/63)
=> H> 1 + 4x(1/8) + 8x (1/16) + (1/17+...+1/63)
=> H> 1+ 1/2 + 1/2 + (1/17+...+1/63)
=> H> 1+1+(1/17+...+1/63)
=> H>1+1
=> H>2
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Cho A = \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}\)
a. Chứng tỏ A > 2.
b. Chứng tỏ a không phải là số tự nhiên.
vì 1/2+1/3+1/4+1/5+1/6+.....+1/11=2,0198765(3)>2 => A>2
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Chứng tỏ rằng :
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}>4\)
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
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Chứng tỏ rằng : \(3< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
Chứng tỏ:
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}>4\)
Chứng tỏ:
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}>4\)
Chứng tỏ rằng
\(3<1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{63}<6\)
So sánh :
Chứng tỏ rằng :
\(1+\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}>4\)
\(=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{32}\right)+\left(\frac{1}{33}+...+\frac{1}{64}\right)\)
\(=1+\frac{1}{2}+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32\)
\(=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=1+\frac{1}{2}.6\)
\(=1+3\)
\(=4\)
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Chứng tỏ rằng:
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{63}+\frac{1}{64}>4\)