当a=0时,则Sn=0.
当a=1时,Sn=1+2+3+…+n=
n(n+1) |
2 |
若a≠0且a≠1时,则Sn=a+2a2+3a3+4a4+…+nan,①
∴aSn=a2+2 a3+3 a4+…+nan+1,②
①-②,得(1-a) Sn=a+a2+a3+…+an-nan+1
=
a(1−an) |
1−a |
∴Sn=
a−an+1 |
(1−a)2 |
nan+1 |
1−a |
若a=0,则Sn=0适合上式.
Sn=
|
∴数列a,2a2,3a2,…,nan的前n项和为
|
n(n+1) |
2 |
a(1−an) |
1−a |
a−an+1 |
(1−a)2 |
nan+1 |
1−a |
|
|