Câu hỏi của Nguyễn Hoàng Gia Bảo

Question

Tính giá trị của biểu thức:

A = $1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}$

Nguyễn Việt Lâm · Accepted Answer

$A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}$

$\Rightarrow2A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2011}}$

$\Rightarrow2A-A=2-\dfrac{1}{2^{2012}}$

$\Rightarrow A=2-\dfrac{1}{2^{2012}}$

Câu trả lời:

$A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}$

$\Rightarrow2A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2011}}$

$\Rightarrow2A-A=2-\dfrac{1}{2^{2012}}$

$\Rightarrow A=2-\dfrac{1}{2^{2012}}$

✪ ω ✪Mùa⚜ hoa⚜ phượng⚜ là⚜ mùa⚜ th... · Answer

$A= 1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+$$\dfrac{1}{2^{2012}}$

⇒$2A=2+1+\dfrac{1}{2}+...+$$\dfrac{1}{2^{2012}}$

⇒$2A-A=(2+1+\dfrac{1}{2}+...+$$\dfrac{1}{2^{2012}}$)$-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2012}}\right)$

⇒$A=2-$$\dfrac{1}{2^{2012}}$

Câu trả lời:

$A= 1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+$$\dfrac{1}{2^{2012}}$

⇒$2A=2+1+\dfrac{1}{2}+...+$$\dfrac{1}{2^{2012}}$

⇒$2A-A=(2+1+\dfrac{1}{2}+...+$$\dfrac{1}{2^{2012}}$)$-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2012}}\right)$

⇒$A=2-$$\dfrac{1}{2^{2012}}$