Question

Tính nhanh

7/2+7/4+7/8+7/16+7/32+......+7/512

Answer 1

#)Giải :

Đặt \(A=\frac{7}{2}+\frac{7}{4}+\frac{7}{8}+\frac{7}{16}+\frac{7}{32}+...+\frac{7}{512}\)

\(A=\frac{7}{2}+\frac{7}{2^2}+\frac{7}{2^3}+\frac{7}{2^4}+...+\frac{7}{2^9}\)

\(\Rightarrow2A=\frac{7}{2^2}+\frac{7}{2^3}+\frac{7}{2^4}+\frac{7}{2^5}+...+\frac{7}{2^{10}}\)

\(\Rightarrow2A-A=\left(\frac{7}{2^2}+\frac{7}{2^3}+\frac{7}{2^4}+\frac{7}{2^5}+...+\frac{7}{2^{10}}\right)-\left(\frac{7}{2}+\frac{7}{2^2}+\frac{7}{2^3}+\frac{7}{2^4}+...+\frac{7}{2^9}\right)\)

\(\Rightarrow A=\frac{7}{2^{10}}-\frac{7}{2}\)

Answer 2

tiếc nhỉ giải theo kiểu học sinh lớp 4 cơ

Answer 3

#)Nếu là cách lớp 4 thì mình chịu @@

Answer 4

hazzz

Answer 5

\(giải:\)

Đặt \(A\)\(=\frac{7}{2}\)\(+\)\(\frac{7}{4}\)\(+\)\(\frac{7}{8}\)\(+\)\(\frac{7}{16}\)\(+\)\(\frac{7}{32}\)\(+\)\(...\)\(+\)\(\frac{7}{512}\)

\(A\)\(=\)\(\frac{7}{2}\)\(+\)\(\frac{7}{2^3}\)\(+\)\(\frac{7}{2^3}\)\(+\)\(\frac{7}{2^4}\)\(+\)\(...\)\(+\)\(\frac{7}{2^9}\)

\(\Rightarrow\)\(2A\)\(-\)\(A\)\(=\)\((\)\(\frac{7}{2^2}\)\(+\)\(\frac{7}{2^3}\)\(+\)\(\frac{7}{2^4}\)\(+\)\(\frac{7}{2^5}\)\(+\)\(...\)\(+\)\(\frac{7}{2^{10}}\)\()\)\(-\)

\((\)\(\frac{7}{2}\)\(+\)\(\frac{7}{2^2}\)\(+\)\(\frac{7}{2^3}\)\(+\)\(\frac{7}{2^4}\)\(+\)\(...\)\(+\)\(\frac{7}{2^9}\)\()\)

\(\Rightarrow\)\(A=\)\(\frac{7}{2^{10}}\)\(-\)\(\frac{7}{2}\)

Answer 6

\(P=\frac{7}{2}+\frac{7}{4}+\frac{7}{8}+\frac{7}{16}+...+\frac{7}{512}\)

\(=\frac{7}{2}+\frac{7}{2^2}+\frac{7}{2^3}+...+\frac{7}{2^9}\)

\(\Rightarrow2P-P=\left(\frac{7}{2^2}+\frac{7}{2^3}+...+\frac{7}{2^{10}}\right)-\left(\frac{7}{2}+\frac{7}{2^2}+...+\frac{7}{2^9}\right)\)

\(\Rightarrow P=\frac{7}{2^{10}}-\frac{7}{2}\)