a) \(A=1-2+2^2-2^3+...+2^{2002}\)
\(\Leftrightarrow\)\(2A=2-2^2+2^3-2^4+....+2^{2003}\)
\(\Leftrightarrow\)\(A+2A=\left(1-2+2^2-2^3+....+2^{2002}\right)+\left(2-2^2+2^3-2^4+...+2^{2003}\right)\)
\(\Leftrightarrow\)\(3A=1+2^{2003}\)
\(\Leftrightarrow\)\(A=\frac{1+2^{2003}}{3}\)
b) bn lm tương tụ nha: lấy 3B + B