#)Giải :
Đặt \(A=\frac{7}{2}+\frac{7}{4}+\frac{7}{8}+\frac{7}{16}+\frac{7}{32}+...+\frac{7}{512}\)
\(A=\frac{7}{2}+\frac{7}{2^2}+\frac{7}{2^3}+\frac{7}{2^4}+...+\frac{7}{2^9}\)
\(\Rightarrow2A=\frac{7}{2^2}+\frac{7}{2^3}+\frac{7}{2^4}+\frac{7}{2^5}+...+\frac{7}{2^{10}}\)
\(\Rightarrow2A-A=\left(\frac{7}{2^2}+\frac{7}{2^3}+\frac{7}{2^4}+\frac{7}{2^5}+...+\frac{7}{2^{10}}\right)-\left(\frac{7}{2}+\frac{7}{2^2}+\frac{7}{2^3}+\frac{7}{2^4}+...+\frac{7}{2^9}\right)\)
\(\Rightarrow A=\frac{7}{2^{10}}-\frac{7}{2}\)
tiếc nhỉ giải theo kiểu học sinh lớp 4 cơ
\(giải:\)
Đặt \(A\)\(=\frac{7}{2}\)\(+\)\(\frac{7}{4}\)\(+\)\(\frac{7}{8}\)\(+\)\(\frac{7}{16}\)\(+\)\(\frac{7}{32}\)\(+\)\(...\)\(+\)\(\frac{7}{512}\)
\(A\)\(=\)\(\frac{7}{2}\)\(+\)\(\frac{7}{2^3}\)\(+\)\(\frac{7}{2^3}\)\(+\)\(\frac{7}{2^4}\)\(+\)\(...\)\(+\)\(\frac{7}{2^9}\)
\(\Rightarrow\)\(2A\)\(-\)\(A\)\(=\)\((\)\(\frac{7}{2^2}\)\(+\)\(\frac{7}{2^3}\)\(+\)\(\frac{7}{2^4}\)\(+\)\(\frac{7}{2^5}\)\(+\)\(...\)\(+\)\(\frac{7}{2^{10}}\)\()\)\(-\)
\((\)\(\frac{7}{2}\)\(+\)\(\frac{7}{2^2}\)\(+\)\(\frac{7}{2^3}\)\(+\)\(\frac{7}{2^4}\)\(+\)\(...\)\(+\)\(\frac{7}{2^9}\)\()\)
\(\Rightarrow\)\(A=\)\(\frac{7}{2^{10}}\)\(-\)\(\frac{7}{2}\)
\(P=\frac{7}{2}+\frac{7}{4}+\frac{7}{8}+\frac{7}{16}+...+\frac{7}{512}\)
\(=\frac{7}{2}+\frac{7}{2^2}+\frac{7}{2^3}+...+\frac{7}{2^9}\)
\(\Rightarrow2P-P=\left(\frac{7}{2^2}+\frac{7}{2^3}+...+\frac{7}{2^{10}}\right)-\left(\frac{7}{2}+\frac{7}{2^2}+...+\frac{7}{2^9}\right)\)
\(\Rightarrow P=\frac{7}{2^{10}}-\frac{7}{2}\)