题目内容
求和:1+2x+3x2+…+nxn-1,x∈R.
解,根据题意,分3种情况讨论:
(1)x=1时,由等差数列前n项和公式可得Sn=1+2+3+…+n=
,
(2)当x≠1时,
设Sn=1+2x+3x2+…+nxn-1,①
则xSn=x+2x2+3x3+…+nxn,②
①-②可得:(1-x)Sn=1+x+x2+…+xn-1-nxn=1-nxn+
则Sn=
.
故当x=0时,Sn=1;
当x=1时,Sn=
,
当x≠0且x≠1时,Sn=
.
(1)x=1时,由等差数列前n项和公式可得Sn=1+2+3+…+n=
n(n+1) |
2 |
(2)当x≠1时,
设Sn=1+2x+3x2+…+nxn-1,①
则xSn=x+2x2+3x3+…+nxn,②
①-②可得:(1-x)Sn=1+x+x2+…+xn-1-nxn=1-nxn+
x(1-xn-1) |
1-x |
则Sn=
1-(n+1)xn+nxn+1 |
(1-x)2 |
故当x=0时,Sn=1;
当x=1时,Sn=
n(n+1) |
2 |
当x≠0且x≠1时,Sn=
1-(n+1)xn+nxn+1 |
(1-x)2 |
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