Question

Tính: a) A=\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+...+\(\dfrac{1}{2^{100}}\)

b) \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{2023.2024}\) 

cứu tôi mng owiiii :((

Answer 1

a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)

\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{101}}\)

\(2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}\)

\(A=1-\dfrac{1}{2^{100}}\)

b) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2023\cdot2024}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)

\(=1-\dfrac{1}{2024}\)

\(=\dfrac{2024}{2024}-\dfrac{1}{2024}\)

\(=\dfrac{2023}{2024}\)

Answer 2

cứu 

Answer 3

okee laaa

Answer 4

   A = \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) + ...+ \(\dfrac{1}{2^{99}}\) + \(\dfrac{1}{2^{100}}\)

2A  = 2.(\(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\)+...+ \(\dfrac{1}{2^{99}}\) + \(\dfrac{1}{2^{100}}\))

2A =  1 + \(\dfrac{1}{2}\)  + \(\dfrac{1}{2^2}\)+.....+ \(\dfrac{1}{2^{98}}\) + \(\dfrac{1}{2^{99}}\)

2A - A  = \(1\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + ...+ \(\dfrac{1}{2^{98}}\) + \(\dfrac{1}{2^{99}}\) - (\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\) +\(\dfrac{1}{2^3}\)+...+\(\dfrac{1}{2^{99}}\)+\(\dfrac{1}{2^{100}}\))

A = 1 + \(\dfrac{1}{2}\)+ \(\dfrac{1}{2^2}\)+ ...+ \(\dfrac{1}{2^{98}}\) + \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\)-...- \(\dfrac{1}{2^{100}}\)

A = 1 - \(\dfrac{1}{2^{100}}\)