Question

tính:

B=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+........+\frac{1}{2^{10}}\)

C=\(\frac{4^{20}+4^{29}}{8^{13}+8^{15}}\)

 

 

 

 

Answer 1

b) \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

\(\Rightarrow B=1-\frac{1}{2^9}\)

c) \(C=\frac{4^{20}+4^{29}}{8^{13}+8^{15}}=\frac{\left(2^2\right)^{20}+\left(2^2\right)^{29}}{\left(2^3\right)^{13}+\left(2^3\right)^{15}}=\frac{2^{40}+2^{58}}{2^{39}+2^{45}}=\frac{2^{40}\left(1+2^{18}\right)}{2^{39}\left(1+2^6\right)}=\frac{2\left(1+2^{18}\right)}{1+2^6}=\frac{2+2^{19}}{1+2^6}\)

Answer 2

\(B=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\\ \Rightarrow2B=1+\frac{1}{2}+....=\frac{1}{2^9}\\ \Rightarrow B=1-\frac{1}{2^{10}}\)

\(C=\frac{4^{20}+4^{29}}{8^{13}+8^{15}}=\frac{\left(2^2\right)^{20}+\left(2^2\right)^{29}}{\left(2^3\right)^{13}+\left(2^3\right)^{15}}=\frac{2^{40}+2^{58}}{2^{39}+2^{45}}=\frac{2^{39}\left(2+2^{19}\right)}{2^{39}\left(1+2^6\right)}\)

Answer 3

\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{10}}\)

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

\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{10}}\right)\)

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

\(\Rightarrow B=\frac{1023}{1024}\)