Câu hỏi của Hải Đăng

Question

Bài 1:

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

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

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

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,

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

c,

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

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

c,

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

Mashiro Shiina · Answer

Đặt:

$X=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}$

$2X=2\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)$

$2X=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}$

$2X-X=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)$$X=\dfrac{1}{2}-\dfrac{1}{2^{2016}}$

$Y=\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{59.61}$

$Y=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{59}-\dfrac{1}{61}$

$Y=\dfrac{1}{5}-\dfrac{1}{61}=\dfrac{56}{305}$

$Z=\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}$

$Z=\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+...+\dfrac{7}{99.101}$

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

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

$Z=\dfrac{7}{2}.\dfrac{100}{101}=\dfrac{700}{202}$Câu trả lời: Đặt: