Chứng minh rằng:
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{63}>2\)
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Chứng minh rằng:
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}>2\)
Chứng minh rằng:
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}>2\)
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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Chứng minh rằng:
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{63}\)
Chứng minh rằng: \(A< 6\)
Ta có:
\(\frac{1}{2}< 6\)
\(\frac{1}{3}< 6\)
\(...\)
\(\frac{1}{63}< 6\)
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{63}< 6\)
\(\Rightarrow A< 6\left(dpcm\right)\)
\(#Jen\)
Trao đổi nếu cần
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nếu làm như bạn thì tổng A < 6 + 6 + 6 +...+ 6 chứ không phải 6 :((((
Chứng minh rằng:
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{63}>2\)
1 tháng 5 2018 lúc 19:37
Chứng minh rằng:
a) \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.........+\frac{1}{100^2}< 2\)
b) \(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.........+\frac{1}{63}< 6\)
Trả lời
a) Đặt \(H=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow H< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Leftrightarrow H< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow H< 1-\frac{1}{100}\)
\(\Leftrightarrow H< \frac{99}{100}\)
\(\Leftrightarrow A< 1+\frac{99}{100}\)
Ta thấy \(\frac{99}{100}< 1\Rightarrow A< 2\)
Vậy A<2 (đpcm)
b) Ta có: 1=1
\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)
\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< \frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)
\(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+...+\frac{1}{15}< \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}=1\)
\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}< \frac{1}{16}+\frac{1}{16}+...+\frac{1}{16}=1\)
\(\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}< \frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}=1\)
\(\Rightarrow B< 1+1+1+1+1+1\)
\(\Rightarrow B< 6\)
Vậy B<6 (đpcm)
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cho:
a) A= 2+\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}+\frac{1}{65}+\frac{1}{66}+\frac{1}{67}\)
chứng minh rằng A>5
b) B= \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{89^2}+\frac{1}{90^2}\)
chứng minh rằng \(\frac{40}{91}\)<B<1
Chứng minh rằng
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{121}-\frac{1}{122}+\frac{1}{123}=\frac{1}{62}+\frac{1}{63}+...+\frac{1}{122}\)-\(\frac{1}{123}\)
Xét \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{123}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{121}+\frac{1}{123}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{122}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{121}+\frac{1}{123}\right)-2\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{61}\right)\)
\(=\frac{1}{62}+\frac{1}{63}+\frac{1}{64}+...+\frac{1}{123}\)
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Bài 1) 4.11.\(\frac{3}{4}.\frac{9}{121}\)
Bài 2) Chứng minh rằng: \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{63}>2\)