Question

求Bn=(2n-8)*(1/3)^(n-1)的前n项和Tn

Answer 1

Bn=(2n-8)*(1/3)^(n-1)

Tn=-6-4*1/3-2*(1/3)^2+0+2*(1/3)^4+...(2n-8)*(1/3)^(n-1)

1/3Tn=-6*1/3-4*(1/3)^2-2*(1/3)^3+0+2*(1/3)^5+...(2n-8)*(1/3)^n

两式相减得

2/3Tn=-6+2*1/3+2*(1/3)2+.2*(1/3)^(n-1)-(2n-8)*(1/3)^n

=-6+2[1/3-(1/3)^n]/(1-1/3)-(2n-8)*(1/3)^n

=-6+1-(1/3)^(n-1)-(2n-8)*(1/3)^n

Tn=-15/2-3/2*[(1/3)^(n-1)+(2n-8)*(1/3)^n]