【答案】
分析:根據(jù)題意,分3種情況討論:(1)x=1時,由等差數(shù)列前n項和公式可得S
n,(3)x≠1時,用錯位相減法,可得答案.
解答:解,根據(jù)題意,分3種情況討論:
(1)x=1時,由等差數(shù)列前n項和公式可得S
n=1+2+3+…+n=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112626074075351/SYS201312021126260740753020_DA/0.png)
,
(2)當(dāng)x≠1時,
設(shè)S
n=1+2x+3x
2+…+nx
n-1,①
則xS
n=x+2x
2+3x
3+…+nx
n,②
①-②可得:(1-x)S
n=1+x+x
2+…+x
n-1-nx
n=1-nx
n+
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112626074075351/SYS201312021126260740753020_DA/1.png)
則S
n=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112626074075351/SYS201312021126260740753020_DA/2.png)
.
故當(dāng)x=0時,S
n=1;
當(dāng)x=1時,S
n=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112626074075351/SYS201312021126260740753020_DA/3.png)
,
當(dāng)x≠0且x≠1時,S
n=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112626074075351/SYS201312021126260740753020_DA/4.png)
.
點評:本題考查數(shù)列的求和,注意按x的值不同,分3種情況討論,容易遺漏x=0的情況.