Chứng minh rằng M < 6 biết:
M = \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{62}+\frac{1}{63}\)
Giúp mk nhé Mai.
Những câu hỏi liên quan
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 tập:
Chứng tỏ rằng:
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}>2\)
Giúp mk nhanh nhé! mk cần gấp ai xong đầu mk sẽ cho 3 tk
1/2+1/3+1/4+...+1/63>1/31+1/31+...+1/31(62 số hạng 1/31)
hay 1/2+1/3+1/4+...+1/63>62 x 1/31
nên 1/2+1/3+1/4+...+1/63>2(dpcm)
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Chứng minh rằng :
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
HELP ME!!!
Các bn giải giúp mik (cách trình bày của lớp 6 nhé)
B < 1+1+1/2.3+1/3.4+...+1/62.63
B < 2+(1/2-1/3+1/3-1/4+...+1/62-1/63)
B < 2+(1/2-1/63)
B < 2+61/126 suy ra B < 2 và 6/126
Mà 2 + 61/126 <6
Suy ra B< 2+6/126<6 suy tiếp B < 6
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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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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: \(3<1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{62}+\frac{1}{63}<6\)
So sánh :
Chứng tỏ rằng :
\(1+\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}>4\)
\(=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{32}\right)+\left(\frac{1}{33}+...+\frac{1}{64}\right)\)
\(=1+\frac{1}{2}+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32\)
\(=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=1+\frac{1}{2}.6\)
\(=1+3\)
\(=4\)
~~ Bố thí cái li.ke ~~
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Chứng minh rằng \(S=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}
Ta có : \(S=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)
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Ta có:
\(S=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\)
\(=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)
Bài toán phụ 1:
Ta có:
1/13<1/12
1/14<1/12
1/15<1/12
=>1/13+1/14+1/15<1/12x3=1/4 (1)
Bài toán phụ 2:
Ta có:
1/61<1/60
1/62<1/60
1/63<1/60
=>1/61+1/62+1/63<1/60x3=1/20 (2)
Từ (1) và (2), ta có:
1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/5+1/4+1/20
1/5+1/13+1/14+1/15+1/61+1/62+1/63<4/20+5/20+1/20
1/5+1/13+1/14+1/15+1/61+1/62+1/63<9/20<1/2
=>1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/2
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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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