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\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+100}\right)\)
\(A=\frac{2}{\left(1+2\right).2:2}.\frac{5}{\left(1+3\right).3:2}.\frac{9}{\left(1+4\right).4:2}...\frac{5049}{\left(1+100\right).100:2}\)
\(A=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{10098}{100.101}\)
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{99.102}{100.101}\)
\(A=\frac{1.2.3...99}{2.3.4...100}.\frac{4.5.6...102}{3.4.5...101}\)
\(A=\frac{1}{100}.\frac{102}{3}=100.34=\frac{1}{100}.34=\frac{17}{50}\)
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)..\left(1-\frac{1}{2000^2}\right)\)
\(=\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}\cdot\frac{3.5}{4^2}\cdot\cdot\cdot\cdot\frac{1998.2000}{1999^2}\cdot\frac{1999.2001}{2000^2}\)
\(=\frac{1}{2}\cdot\frac{2001}{2000}=\frac{2001}{4000}\)
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{1999^2}\right)\left(1-\frac{1}{2000^2}\right)\)
=\(\left(\frac{4}{4}-\frac{1}{4}\right)\left(\frac{9}{9}-\frac{1}{9}\right)...\left(\frac{3996001}{3996001}-\frac{1}{3996001}\right)\left(\frac{4000000}{4000000}-\frac{1}{4000000}\right)\)
=\(\frac{3}{4}.\frac{8}{9}....\frac{3996000}{3996001}.\frac{3999999}{4000000}\)
=\(\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{1998.2000}{1999.1999}.\frac{1999.2001}{2000.2000}\)
=\(\frac{1.3.2.4.3.6.....1998.2000.1999.2001}{2.2.3.3.4.4....1999.1999.2000.2000}=\frac{1.2001}{2.2000}=\frac{2001}{4000}\)
a)Ta có:
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{98^2}-1\right)\left(\frac{1}{99^2}-1\right)\)
\(=\left(\frac{1}{2.2}-1\right)\left(\frac{1}{3.3}-1\right)\left(\frac{1}{4.4}-1\right)....\left(\frac{1}{98.98}-1\right)\left(\frac{1}{99.99}-1\right)\)
\(=\left(-\frac{3}{2.2}\right).\left(-\frac{8}{3.3}\right).\left(-\frac{15}{4.4}\right)...\left(-\frac{9603}{98.98}\right).\left(-\frac{9800}{99.99}\right)\)
\(=\left[\left(-1\right).\left(-1\right).\left(-1\right)...\left(-1\right)\right].\frac{3}{2.2}.\frac{8}{3.3}.\frac{15}{4.4}...\frac{9603}{98.98}.\frac{9800}{99.99}\)
|------------------------98 số -1--------------------|
\(=\left(-1\right)^{98}.\frac{1.3}{2.3}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{95.97}{98.98}.\frac{98.100}{99.99}\)
\(=\frac{1.3}{2.3}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{95.97}{98.98}.\frac{98.100}{99.99}\)
\(=\frac{1.3.2.4.3.5...95.97.98.100}{2.2.3.3.4.4...98.98.99.99}\)
Ta sẽ rút gọn các thừa số chung ở tử và mẫu
\(=\frac{1.100}{2.99.99}\)
\(=\frac{50}{9801}\)
Vậy \(A=\frac{50}{9801}\)
cho mik hỏi bước 3 chỗ \(\frac{3}{2.2}\)sai o duoi lai la\(\frac{3}{2.3}\)vay
\(\Rightarrow2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2014}\)
\(\Rightarrow2A-A=A=1-\left(\frac{1}{2}\right)^{2015}\)
Với B tương tự nhưng là lấy 3B
Ta có:\(\left(x-1\right)\left(x+1\right)=x\left(x-1\right)+x-1^2=x^2-x+x-1=x^2-1\)
Áp dụng:\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot...\cdot\frac{2014^2-1}{2014\cdot2014}\)
\(=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot...\cdot\frac{2013\cdot2015}{2014^2}\)
\(=\frac{1}{2}\cdot\frac{2015}{2014}=\frac{2015}{4028}\)