∵Sn=1+3x+5x2+7x3+…+(2n-1)xn-1,
∴xSn=x+3x2+…+(2n−3)xn−1+(2n-1)xn,
两式相减得(1-x)Sn=1+2x+2x2+…+2xn-1-(2n-1)xn,
①当x≠1,0时,由等比数列的求和公式得:(1-x)Sn=-1+
2(1−xn) |
1−x |
∴Sn=
(2n−1)xn+1−(2n+1)xn+(1+x) |
(1−x)2 |
②当x=1时,Sn=1+3+5+…+(2n-1)=
n(1+2n−1) |
2 |
③当x=0时,Sn=1+0=1.
2(1−xn) |
1−x |
(2n−1)xn+1−(2n+1)xn+(1+x) |
(1−x)2 |
n(1+2n−1) |
2 |