Câu hỏi của Phạm Thị Thạch Thảo

Question

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}+...+\dfr...

Mới vô · Accepted Answer

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,

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

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) $\dfrac{7}{3}+\dfrac{7}{15}+.........+\dfrac{7}{9999}$

$=\dfrac{7}{1.3}+\dfrac{7}{3.5}+...........+\dfrac{7}{99.101}$

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

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

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

$=\dfrac{7}{2}.\dfrac{100}{101}=\dfrac{350}{101}$Câu trả lời: a) Đặt :