错位相减法 n·an=n·(-1/2)^﹙n-1﹚,求﹛nan﹜的前n项和求sn-(-1/2)sn以后的步骤及_数学解答

错位相减法 n·an=n·(-1/2)^﹙n-1﹚,求﹛nan﹜的前n项和
求sn-(-1/2)sn以后的步骤及

人气：274 ℃　时间：2020-09-19 10:55:11

解答

Sn=1+2×(-1/2)+3×(-1/2)²+…+n·(-1/2)^(n-1),则
(-1/2)Sn=1×(-1/2)+2×(-1/2)²+…+(n-1)·(-1/2)^(n-1)+n·(-1/2)^n,
两式相减,得
Sn -(-1/2)Sn=1+(-1/2)+(-1/2)²+…+(-1/2)^(n-1)-n·(-1/2)^n,
即(3/2)Sn=[1-(-1/2)^n]/(1+1/2)-n·(-1/2)^n,
所以,Sn=(4/9)[1-(-1/2)^n]-(2/3)n·(-1/2)^n.