Cho C = \(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{99\times100}\). so sánh c với \(\frac{1}{2}\)
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Cho C = \(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{99\times100}\). So sanh C với \(\frac{1}{2}\)
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\Rightarrow C>\frac{1}{2}\)
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Ta có : \(C=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy \(C< \frac{1}{2}\)
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\(\frac{1\times2}{2\times3}+\frac{2\times3}{3\times4}+\frac{3\times4}{4\times5}+...+\frac{98\times99}{99\times100}\)
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
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a,A=\(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
b,B=\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{998\times999\times100}\)
c,C=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1\times98+2\times97+3\times96+...+98\times1}\)
\(\left(1-\frac{2}{2\times3}\right)\left(1-\frac{2}{3\times4}\right)\left(1-\frac{2}{4\times5}\right)...\left(1-\frac{2}{99\times100}\right)\)= ?
Tính:\(\left(1-\frac{2}{2\times3}\right)\left(1-\frac{2}{3\times4}\right)\left(1-\frac{2}{4\times5}\right)....\left(1-\frac{2}{99\times100}\right)\)
Tính nhanh:
C=\(\frac{3}{2\times3\times4}+\frac{3}{3\times4\times5}+\frac{3}{4\times5\times6}+......+\frac{3}{98\times99\times100}\)
C = \(\frac{3}{2.3.4}+\frac{3}{3.4.5}+.....+\frac{3}{98.99.100}\)
C = \(3.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
C = \(3.\frac{1}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+...+\frac{100}{98.99.100}-\frac{99}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\frac{1649}{9900}\)
C = \(\frac{1649}{6600}\)
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Hồ Thu Giang cần chứ nếu được cảm ơn nha
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Cho A=\(\frac{1\times2-1}{2!}+\frac{2\times3-1}{3!}+\frac{3\times4-1}{4!}+.....+\frac{99\times100-1}{100!}<2\)
Bài 4 : Tính nhanh :
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+....+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
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\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
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Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
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Xem thêm câu trả lời
1\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+............+\frac{1}{99\times100}+\frac{1}{100\times101}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}.\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
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