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Phạm Thị Thạch Thảo

Tính tổng:

a,\(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)

b,\(\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{59.61}\)

c,\(\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\)

Giúp mk vs nha@Hồng Phúc Nguyễn @Mới vô @Bùi Thị Vân

Mới vô
9 tháng 8 2017 lúc 15:55

a,

\(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\\ =1\cdot\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\\ =\left(2-1\right)\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\\ =\left(2-1\right)\cdot\dfrac{1}{2^2}+\left(2-1\right)\cdot\dfrac{1}{2^3}+...+\left(2-1\right)\cdot\dfrac{1}{2^{2006}}\\ =\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{2005}}-\dfrac{1}{2^{2006}}\\ =\dfrac{1}{2}-\dfrac{1}{2^{2006}}\\ =\dfrac{2^{2005}}{2^{2006}}-\dfrac{1}{2^{2006}}\\ =\dfrac{2^{2005}-1}{2^{2006}}\)

b,

\(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{59\cdot61}\\ =\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{59}-\dfrac{1}{61}\\ =\dfrac{1}{5}-\dfrac{1}{61}\\=\dfrac{56}{305}\)

c,

\(\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\\ =\dfrac{7}{2}\cdot\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{9999}\right)\\ =\dfrac{7}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{7}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{7}{2}\cdot\left(1-\dfrac{1}{101}\right)\\ =\dfrac{7}{2}\cdot\dfrac{100}{101}\\ =\dfrac{350}{101}\)

Nguyễn Thanh Hằng
9 tháng 8 2017 lúc 15:58

a) Đặt :

\(A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+............+\dfrac{1}{2^{2006}}\)

\(\Leftrightarrow2A=1+\dfrac{1}{2^2}+...........+\dfrac{1}{2^{2005}}\)

\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2^2}+......+\dfrac{1}{2^{2005}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+.......+\dfrac{1}{2^{2006}}\right)\)

\(\Leftrightarrow A=1-\dfrac{1}{2^{2006}}\)

b) \(\dfrac{2}{5.7}+\dfrac{2}{7.9}+.........+\dfrac{2}{59.61}\)

\(=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+.........+\dfrac{1}{59}-\dfrac{1}{61}\)

\(=\dfrac{1}{5}-\dfrac{1}{61}\)

\(=\dfrac{56}{305}\)

c) \(\dfrac{7}{3}+\dfrac{7}{15}+.........+\dfrac{7}{9999}\)

\(=\dfrac{7}{1.3}+\dfrac{7}{3.5}+...........+\dfrac{7}{99.101}\)

\(=\dfrac{7}{2}\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+..........+\dfrac{1}{99.101}\right)\)

\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{7}{2}\left(1-\dfrac{1}{101}\right)\)

\(=\dfrac{7}{2}.\dfrac{100}{101}=\dfrac{350}{101}\)


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