Question

求等比数列(2n-1)*2^n的前n项和

Answer 1

S=1*2^1+3*2^2+ …… + (2n-1)*2^n (1)

2S= 1*2^2+3*2^3+……+(2n-3)*2^n+(2n-1)*2^(n+1) (2)

(1)-(2)得-S=1*2^1+2*(2^2+2^3……+2^n)-(2n-1)*2^(n+1)

=2*(2^1+2^2+2^3……+2^n)-(2n-1)*2^(n+1) -2^1

=2*2*(2^n-1)-(2n-1)*2^(n+1) -2^1

=-(2n-3)*2^(n+1)-6

∴等比数列(2n-1)*2^n的前n项和S=(2n-3)*2^(n+1)+6