CTR
B = \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}<6\)
Những câu hỏi liên quan
B=\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
C=\(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
Afrac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{63}2 C1+frac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{63} 6B1+frac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{100^2} 2 Dfrac{1}{2}.frac{3}{4}.frac{5}{6}....frac{9999}{10000} frac{1}{100}Mọi người giúp mik nhé, mik đang ôn thi nên cần gấp!
Đọc tiếp
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}>2\) \(C=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
\(B=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\) \(D=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}< \frac{1}{100}\)
Mọi người giúp mik nhé, mik đang ôn thi nên cần gấp!
chứng minh:
B=\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
C=\(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
b, Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
..................
\(\frac{1}{100^2}< \frac{1}{99.100}\)
Nên C < \(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}\)
<=> C < \(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\)
<=> C < \(1+1-\frac{1}{100}\)
<=> C < \(2-\frac{1}{100}=\frac{199}{100}\)
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\(B=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+...+\left(\frac{1}{2^5}+...+\frac{1}{2^6-1}\right)\)
\(B< 1+\frac{1}{2}.2+\frac{1}{4}.4+...+\frac{1}{2^5}.32\)
\(B< 1+1+1+...+1\)( 6 số 1)
B<1.6=6
\(C=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
\(C< 1+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.10}\right)=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)\(=1+\left(1-\frac{1}{100}\right)< 1+1=2\)
Vậy C<2
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C<1+1/1.2+1/2.3+...+1/99.100
=>C<1+1/1-1/100
<=>C<1+99/100
=>C<199/200
=>C<2
BẠN TỰ GIẢI CHI TIẾT HƠN NHA ^_^
MÌNH GIẢI HƠI TẮT CHÚT :))
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\(CM:\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}>2\)
\(CM:\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}>2\)
CHỨNG MINH:
\(\frac{ }{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{63}+\frac{1}{64}>3}\)
21 tháng 4 2019 lúc 21:03
a) Chứng minh: \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{60}< \frac{3}{2}\)
b) Chứng minh: \(3< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
Chung minh rang : 1+\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}+\frac{1}{64}>4\)
Ta có:\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..........+\frac{1}{64}\)
=\(1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+.........+\left(\frac{1}{33}+......+\frac{1}{64}\right)\)
\(>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+...+\left(\frac{1}{64}+\frac{1}{64}+.........+\frac{1}{64}\right)\)
=\(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
=4
Vậy \(1+\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{64}>4\)
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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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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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