cho C = 1/1.2 + 1/3.4 + 1/5.6 + . . . + 1/97.98 + 1/99.100 & D = 1/51.100 + 1/52.99 + 1/53.98 + . . . + 1/99.52 + 1/100.51
Chung minh C : D ko nhan gia tri la mot so tu nhien
Những câu hỏi liên quan
Cho a/b = 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/97.98 + 1/99.100
Chứng minh rằng a chia hết cho 151
A=1/1.2+1/3.4+1/5.6+....+1/97.98+1/99.100 B=1/50+1/51+1/52+....+1/99+1/100 Tính A-B
Tinh tong
F=1.2+3.4+5.6+...+97.98+99.100
G=1^2+2^2+3^2+....+99^2+100^2
G = \(1^2\)+\(2^2\)+ \(3^2\)+....+\(100^2\)
G=1 +2(1+1) +3(2+1) +..... + 100(99+1)
G=1 + 1.2+ 2 + 2.3 +3+ ......+ 99.100+100
G=(1+2+3+....+100) +(1.2+2.3+.....+99.100)
G= \(\frac{100\left(100+1\right)}{2}\)+\(\frac{100\left(100-1\right)\left(100+1\right)}{3}\)
G=5050+333300
G=338350
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CHO : \(M=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.......+\frac{1}{97.98}+\frac{1}{99.100}\)
Chứng Minh: \(\frac{7}{12}< M< \frac{5}{6}\)
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tham khảo ở đó nhé!!!
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CHO :A= \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+..........+\dfrac{1}{97.98}+\dfrac{1}{99.100}\)
CHỨNG MINH :\(\dfrac{7}{12}< A< \dfrac{5}{6}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{97.98}+\dfrac{1}{99.100}\)
\(A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{9506}+\dfrac{1}{9900}\)
\(A=\left(\dfrac{1}{2}+\dfrac{1}{12}\right)+\left(\dfrac{1}{30}+...+\dfrac{1}{9506}+\dfrac{1}{9900}\right)\)
\(A>\dfrac{1}{2}+\dfrac{1}{12}\Rightarrow A>\dfrac{7}{12}\left(1\right)\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{97.98}+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{5}{6}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< \dfrac{5}{6}\left(2\right)\)
\(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)
\(\rightarrowđpcm\)
Chúc bạn học tốt!
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1, CMR: \(\frac{7}{12}<\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{97.98}+\frac{1}{99.100}<\frac{5}{6}\)
2, CMR: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{47.48}+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
Tính tổng
F=1.2+3.4+5.6+6.7+........+97.98+99.100
Fx3=1x2x3+2x3x(4-1)+3x4x(5-2)+4x5x(6-3)+...+99x100x(101-98
Fx3=1x2x3+2x3x4-1x2x3+3x4x5-2x3x4+4x5x6-3x4x5+...+99x100x101-98x99x100
Fx3=99x100x101
F=333300
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Tính tổng
F=1.2+3.4+5.6+7.8+....+97.98+99.100
3F = 1 . 2 . 3 + 3 . 4 . ( 5 - 2 ) + 5 . 6 . ( 7 - 4 ) +.....+ 99 . 100 . (101 - 98 )
3F = 1. 2 . 3 + 3. 4 . 5 - 2 . 3 . 4 + 5 . 6 . 7 - 4 . 5 . 6 +.....+ 99 . 100 . 101 - 98 . 99 . 100
3F = 1 . 2 . 3 + 99 . 100. 101
3F = 3 . 2 + 3 . 33 . 100 . 101
3F = 3 ( 2 + 333 300)
=>F = 3 . 333 302 : 3
=> F = 333 302
Vậy F = 333 302
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tính A-B biết
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{97.98}+\frac{1}{99.100}\)
\(B=\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
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Nhầm tưởng tính tích :v
Ta có :
\(B=\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}+\frac{1}{100}< \frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}=50.\frac{1}{51}=\frac{50}{51}< \frac{99}{100}\)
\(\Leftrightarrow A>B\)
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~ Rim Ceil ~:Chuyên Quốc Học ở đâu thì ko biết nhưng bài như thế này mak làm sai~
\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{100}\right)\)
\(=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100}\right)\)
\(=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+....+\frac{1}{99}+\frac{1}{100}\)
\(\Rightarrow A-B=0\)
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