Câu hỏi của Chu Phan Diệu Thảo

Question

Tính: $-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-...-\frac{1}{1024}$

Dương Kim Chi · Accepted Answer

Ta có: $-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)$đặt $A=1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}$   $\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}$$A-\frac{1}{2}A=\frac{1}{2}A\Rightarrow A=\frac{1-\frac{1}{2^{11}}}{\frac{1}{2}}=2-\frac{1}{2^{10}}$Câu trả lời: Ta có: $-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)$đặt $A=1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}$   $\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}$$A-\frac{1}{2}A=\frac{1}{2}A\Rightarrow A=\frac{1-\frac{1}{2^{11}}}{\frac{1}{2}}=2-\frac{1}{2^{10}}$

Trần Minh Hoàng · Answer

$-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}$$=-1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)$Đặt  $A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}$$2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}$$2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)$$A=1-\frac{1}{1024}=\frac{1023}{1024}$Vậy, $-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}=-1-A=-1-\frac{1023}{1024}=-\frac{2047}{1024}$Câu trả lời: $-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}$$=-1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)$Đặt  $A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}$$2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}$$2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)$$A=1-\frac{1}{1024}=\frac{1023}{1024}$Vậy, $-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}=-1-A=-1-\frac{1023}{1024}=-\frac{2047}{1024}$

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