\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(2A+A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3B-B=1-\frac{1}{3^{99}}\)
\(B=\frac{1-\frac{1}{3^{99}}}{2}\)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(2A+A=\left(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-^2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\right)\)
\(3A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
Chúc bạn học tốt ~
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2B=1-\frac{1}{3^{99}}\)
\(B=\frac{1-\frac{1}{3^{99}}}{2}\)
Chúc bạn học tốt ~