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1/2 . P = 1/2.6 + 1/6.10 + 1/10.14 + ... + 1/198.202
4.1/2. P= 4/2.6 + 4/6.10 + 4/10.14 + ... + 4/198.202
2P=1/2-1/6+1/6-1/10+1/10-1/14+...+1/198-1/202
2P=1/2-1/202=50/101
P=50/101:2=50/101.1/2=25/101
\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{14.9}+...+\frac{1}{198.101}\)
\(A=\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
Ta thấy : thừa số thứ nhất ở mẫu của phân số liền sau = thừa số thứ nhất của phân số liền trước + 4
Thừa số thứ hai ở mẫu của phân số liền sau = thừa số thứ hai của phân số liền trước + 2
\(4A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(4A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{101-99}{99.101}\)
4A= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101.4}=\frac{25}{101}\)
\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)
\(A=2\times\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{198.202}\right)\)
\(A=2\times\frac{1}{4}\times\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{198}-\frac{1}{202}\right)\)
\(A=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{202}\right)\)
\(A=\frac{1}{2}\times\frac{50}{101}\)
\(A=\frac{25}{101}\)
\(A=\frac{1}{2\cdot3}+\frac{1}{6\cdot5}+\frac{1}{10\cdot7}+\frac{1}{14\cdot9}+...+\frac{1}{202\cdot103}\)
\(A=\frac{2}{2\cdot6}+\frac{2}{6\cdot10}+\frac{2}{10\cdot14}+\frac{2}{14\cdot18}+...+\frac{2}{202\cdot206}\)
\(A\cdot2=2\left(\frac{2}{2\cdot6}+\frac{2}{6\cdot10}+\frac{2}{10\cdot14}+\frac{2}{14\cdot18}+...+\frac{2}{202\cdot206}\right)\)
\(A\cdot2=\frac{4}{2\cdot6}+\frac{4}{6\cdot10}+\frac{4}{10\cdot14}+\frac{4}{14\cdot18}+...+\frac{4}{202\cdot206}\)
\(A\cdot2=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{202}-\frac{1}{206}\)
\(A\cdot2=\frac{1}{2}-\frac{1}{206}\)
\(A\cdot2=\frac{103}{206}-\frac{1}{206}\)
\(A\cdot2=\frac{51}{103}\)
\(A\cdot2=\frac{51}{103}\div2=\frac{51}{206}\)
\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)
\(A=2.\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{198.202}\right)\)
\(A=2.\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{198}-\frac{1}{202}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)
\(A=\frac{1}{2}.\frac{50}{201}\)
\(A=\frac{25}{101}\)
\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)
\(A=2.\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{198.202}\right)\)
\(A=2.\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{198}-\frac{1}{202}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)
\(A=\frac{1}{2}.\frac{50}{201}\)
\(A=\frac{25}{101}\)
Như bạn kia là rất đúng
Đặt \(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}.\)
\(2A=\frac{2}{2.3}+\frac{2}{6.5}+\frac{2}{10.7}+...+\frac{2}{198.101}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(4A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)\(=\frac{100}{101}\)
\(A=\frac{100}{101}\div4=\frac{25}{101}\)
\(\Rightarrow\frac{25}{101}x=1\)
\(x=1\div\frac{25}{101}=\frac{101}{25}\)
\(A=\frac{1}{2.1.3}+\frac{1}{2.3.5}+\frac{1}{2.5.7}+...+\frac{1}{2.99.101}\)
\(=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{100}{101}\)
\(=\frac{50}{101}\)
\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)
Đặt :\(M=\frac{1}{2.6}+\frac{1}{6.10}+...+\frac{1}{194.198}\)
\(M=\frac{1}{4}\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{194}-\frac{1}{198}\right)\)
\(M=\frac{1}{4}\left(\frac{1}{2}-\frac{1}{198}\right)\)
\(M=\frac{1}{4}.\frac{49}{99}\)
\(M=\frac{49}{396}\)
Đặt \(N=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(N=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(N=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(N=\frac{1}{2}.\frac{98}{303}\)
\(N=\frac{49}{303}\)
Vậy ta có : A = M + N = \(\frac{49}{396}+\frac{49}{303}\) , bạn tự tính luôn nha