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Ta có : \(B=\frac{1}{2}-\frac{1}{2^2}+...-\frac{1}{2^{100}}\)
\(\Rightarrow2B=1-\frac{1}{2}+\frac{1}{2^2}-...-\frac{1}{2^{99}}\)
\(\Rightarrow2B+B=\left(1-\frac{1}{2}+\frac{1}{2^2}-...-\frac{1}{2^{99}}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+...-\frac{1}{2^{100}}\right)\)
\(\Rightarrow3B=1-\frac{1}{2}+\frac{1}{2^2}-...-\frac{1}{2^{99}}+\frac{1}{2}-\frac{1}{2^2}+...-\frac{1}{2^{100}}\)
\(\Rightarrow3B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow3B=1-\frac{1}{2^{100}}\)
\(\Rightarrow B=\frac{1-\frac{1}{2^{100}}}{3}\)
A = \(\frac{1}{2}\)\(-\)\(\frac{1}{2^2}\)\(+\)\(\frac{1}{2^3}\)\(-\)\(\frac{1}{2^4}\)\(+\)........... \(+\)\(\frac{1}{2^{99}}\)\(-\)\(\frac{1}{2^{100}}\)
2A = 1 - \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)- \(\frac{1}{2^3}\)+.........+ \(\frac{1}{2^{98}}\)- \(\frac{1}{2^{99}}\)
2A + A =( 1 - \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)- \(\frac{1}{2^3}\)+.........+ \(\frac{1}{2^{98}}\)- \(\frac{1}{2^{99}}\)) \(+\)( \(\frac{1}{2}\)\(-\)\(\frac{1}{2^2}\)\(+\)\(\frac{1}{2^3}\)\(-\)\(\frac{1}{2^4}\)\(+\)........... \(+\)\(\frac{1}{2^{99}}\)\(-\)\(\frac{1}{2^{100}}\))
3A = 1 \(-\) \(\frac{1}{2^{100}}\)
\(\Rightarrow\)A = \(\frac{1-\frac{1}{2^{100}}}{3}\)= \(\frac{1}{3}\)
\(M=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(2M=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)
\(2M+M=\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^{98}}-\frac{1}{2^{99}}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\right)\)
\(3M=1-\frac{1}{2^{100}}\)
\(M=\frac{1-\frac{1}{2^{100}}}{3}\)