Question

1/2+1/4+1/8+1/16+...+1/1024

Answer 1

Đặt $A=\dfrac12+\dfrac14+\dfrac18+\dfrac{1}{16}+...+\dfrac{1}{1024}$

$A=\dfrac12+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}$

$\dfrac12\cdot A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{11}}$

$A-\dfrac{1}{2}A=(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}})-(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{11}})$

$\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{11}}$

$\dfrac{1}{2}A=\dfrac{1}{2}\cdot(1-\dfrac{1}{2^{10}})$

$\Rightarrow A=1-\dfrac{1}{2^{10}}$

Vậy: ...

$Toru$