Tham khảo:Câu hỏi của Nguyễn Thị Thanh Bình - Toán lớp 7 - Học trực tuyến OLM

512-\(\frac{512}{2}\)-\(\frac{512}{2^2}\)-\(\frac{512}{2^3}\)-....-\(\frac{512}{2^{10}}\)

=512-256-\(\frac{2^9}{2^2}\)-\(\frac{2^9}{2^3}\)-\(\frac{2^9}{2^4}\)-\(\frac{2^9}{2^5}\)-\(\frac{2^9}{2^6}\)-\(\frac{2^9}{2^7}\)-\(\frac{2^9}{2^8}\)-\(\frac{2^9}{2^9}\)-\(\frac{2^9}{2^{10}}\)

=512-256-128-64-32-16-8-4-2-\(\frac{1}{2}\)

=\(\frac{3}{2}\)

Đặt \(Q=512-\frac{512}{2}-\frac{512}{2^2}-...-\frac{512}{2^{10}}\)

 \(=512-512\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

Đặt  A là tên biểu thức trong ngoặc ta cs:

\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(A=1-\frac{1}{2^{10}}\)

Thay A vào Q ta được:

\(Q=512-512\left(1-\frac{1}{2^{10}}\right)=512-512+\frac{512}{2^{10}}=\frac{2^9}{2^{10}}=\frac{1}{2}\)

\(M=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\)

\(M=512-\frac{512}{2}-\frac{512}{4}-\frac{512}{8}-...-\frac{512}{1024}\)

\(M=\frac{1024}{2}-\frac{512}{2}-\frac{256}{2}-\frac{128}{2}-...-\frac{1}{2}\)

\(M=\frac{1024}{2}-\left(\frac{512}{2}+\frac{256}{2}+\frac{128}{2}+\frac{64}{2}+...+\frac{1}{2}\right)\)

\(M=\frac{1024}{2}-\frac{1023}{2}\)

\(M=\frac{1}{2}\)

\(M=0,5\)

\(M=512-\frac{512}{2^2}-....-\frac{512}{2^{10}}\)
\(=2^9-\frac{2^9}{2}-.....-\frac{2^9}{10}\)
\(=2^9-2^8-....-\frac{1}{2}\)
\(2M=2^{10}-2^9-....-1\)
\(M=\left(2^{10}-...-1\right)-2^9+2^8+....+1+\frac{1}{2}\)
\(M=2^{10}-2.2^9+\frac{1}{2}\)
\(M=\frac{1}{2}\)

ủa sao ông tự hỏi tự trả lời

\(\Rightarrow\frac{M}{512}=1-\frac{1}{2}-\frac{1}{2^2}-.....-\frac{1}{2^{10}}\)

\(\Rightarrow2.\left(\frac{M}{512}\right)=2-1-\frac{1}{2}-.....-\frac{1}{2^9}\)

\(\Rightarrow2.\left(\frac{M}{512}\right)-\frac{M}{512}=\left(2-1-\frac{1}{2}-.....-\frac{1}{2^9}\right)-\left(1-\frac{1}{2}-\frac{1}{2^2}-.....-\frac{1}{2^{10}}\right)\)

\(\Rightarrow\frac{M}{512}=-\frac{1}{2^{10}}\)

\(\Rightarrow M=-\frac{1}{2}\)