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a) Sửa tí: \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)
Đặt \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)
\(\Rightarrow2A=2.\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\)
\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)
\(\Rightarrow2A-A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2006}}\right)\)
\(\Rightarrow A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}-1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{2006}}\)
\(\Rightarrow A=2-\dfrac{1}{2^{2006}}\)
b) Đặt \(A=\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{1}{50.61}\)
\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(A=\dfrac{1}{5}-\dfrac{1}{61}\)
\(A=\dfrac{56}{305}\)
c) Đặt \(A=\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\)
\(A=\dfrac{7}{2}.2.\left(\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{9999}\right)\)
\(A=\dfrac{7}{2}.\left(1-\dfrac{1}{101}\right)\)
\(A=\dfrac{7}{2}.\dfrac{100}{101}\)
\(A=\dfrac{256}{101}\)
A=3/4.(1/5.7+1/7.9+....+1/59.61)
A=3/4.(1/5-1/7+1/7-1/9+...+1/59-1/61)
A=3/4.(1/5-1/61)
A=3/4.56/305
A=42/305
mình làm cho bạn phần A thôi nhé còn phần B mình chưa nghĩ ra cách làm ahihi!
\(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\)
\(=3\left(\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+...+\dfrac{1}{59\cdot61}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\) \(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)
A=\(\dfrac{3}{5.7}\)+\(\dfrac{3}{7.9}\)+...+\(\dfrac{3}{59.61}\)
A=\(\dfrac{3}{2}\)(\(\dfrac{2}{5.7}\)+\(\dfrac{2}{7.9}\)+...+\(\dfrac{2}{59.61}\))
A=\(\dfrac{3}{2}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{9}\)+..+\(\dfrac{1}{59}\)-\(\dfrac{1}{61}\))
A=\(\dfrac{3}{2}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{61}\))
A=\(\dfrac{3}{2}\)(\(\dfrac{61-5}{5.61}\))
A=\(\dfrac{3}{2}\).\(\dfrac{56}{305}\)
A=\(\dfrac{84}{305}\)
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
Theo quy luật thì mình nghĩ đáng lẽ \(\dfrac{4}{5.9}\)phải là\(\dfrac{4}{7.9}\)Bạn có chép sai đề ko?
A=1-\(\dfrac{4}{5.7}-\dfrac{4}{7.9}-\dfrac{4}{9.11}...-\dfrac{4}{59.61}\)
A=\(1-\left(\dfrac{4}{5.7}+\dfrac{4}{7.9}+\dfrac{4}{9.11}+...+\dfrac{4}{59.61}\right)\)
Đặt B=\(\dfrac{4}{5.7}+\dfrac{4}{7.9}+\dfrac{4}{9.11}+...+\dfrac{4}{59.61}\)
B=\(2\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{59.61}\right)\) B=\(2\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\) B=\(2\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=2.\dfrac{56}{305}\) B=\(\dfrac{112}{305}\) \(\Rightarrow A=1-\dfrac{112}{305}=\dfrac{193}{305}\)
\(T=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{59\cdot61}\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{59\cdot61}\right)\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{59.61}\)
\(=3.\left(\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{59.61}\right)\)
\(=3.\dfrac{1}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}.\dfrac{56}{305}\)
\(=\dfrac{84}{305}\)