\(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+....+\left(1-\frac{1}{512}\right)+\left(1-\frac{1}{1024}\right)\)
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\(A=\left(1-\frac{1}{2^1}\right)+\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^3}\right)+...+\left(1-\frac{1}{2^9}\right)+\left(1-\frac{1}{2^{10}}\right)\)
\(A=\left(1+1+1+...+1+1\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
10 số 1
\(A=10-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)
\(2B-B=1-\frac{1}{2^{10}}=B\)
=> \(A=10-\left(1-\frac{1}{2^{10}}\right)\)
=> \(A=10-1+\frac{1}{2^{10}}\)
=> \(A=9\frac{1}{1024}\)
\(C=\left(\frac{1}{2}-1\right)+\left(1-\frac{3}{4}\right)+\left(\frac{7}{8}-1\right)+...+\left(1-\frac{1023}{1024}\right)\)
\(C=\left(\frac{1}{2^1}-\frac{2}{2}\right)+\left(\frac{2^2}{2^2}-\frac{3}{2^2}\right)+...+\left(\frac{1024}{1024}-\frac{1023}{2^{10}}\right)\)
\(C=\frac{-1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2C=-1+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2C+C=\left(-1+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{9}\right)+\left(-\frac{1}{2}+\frac{1}{2^2}-..+\frac{1}{2^{10}}\right)\)
\(3C=\frac{1}{2^{10}}-1\)
\(C=\frac{\frac{1}{2^{10}}-1}{3}\)
hok tốt!!
\(a)\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)\(=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{3}{10}.\frac{-4}{3}}=\frac{5}{24}\)
Hok tốt