1+3×2+5×2²+7×2³+.求前n项之和

设A(n) = 1 + 3×2 + 5×2² + 7×2³ + ...+ (2n-1)*2^(n-1)
2A(n) = 2 + 3×2² + 5×2³ + ...+ (2n-3)*2^(n-1) + (2n-1)*2^n
两式相减,得
A(n) = -1 - 2*2 - 2*2^2 - 2*2^3 - ...- 2*2^(n-1) + (2n-1)*2^n = -1 - (2^(n+1) - 4) + (2n-1)*2^n
= (2n-3)*2^n + 3