求数列(2n-3/2^(n-3))的前n项和

Question

求数列(2n-3/2^(n-3))的前n项和

Answer 1

设数列(2n-3/2^(n-3))的前n项和为Sn
则Sn=-4+2+3+5/2+7/2^2+9/2^3+.+(2n-5)/2^(n-4)+(2n-3)/2^(n-3)
=1+5/2+7/2^2+9/2^3+.+(2n-5)/2^(n-4)+(2n-3)/2^(n-3).(1)
把(1)式两端同乘以(1/2),得
Sn/2=1/2+5/2^2+7/2^3+9/2^4+.+(2n-5)/2^(n-3)+(2n-3)/2^(n-2).(2)
(1)-(2),得
Sn/2=1/2+5/2+2/2^2+2/2^3+.+2/2^(n-3)-(2n-3)/2^(n-2)
=2+1+1/2+1/2^2+.+1/2^(n-4)-(2n-3)/2^(n-2)
=2+(1-1/2^(n-3))/(1-1/2)-(2n-3)/2^(n-2)
=2+2-1/2^(n-4)-(2n-3)/2^(n-2)
=4-(2n+1)/2^(n-2)
故Sn=8-(2n+1)/2^(n-3).