设a为常数,求数列a,2a2,3a2,…,nan的前n项和.
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设a为常数,求数列a,2a2,3a2,…,nan的前n项和.
![设a为常数,求数列a,2a2,3a2,…,nan的前n项和.](/uploads/image/z/1023525-45-5.jpg?t=%E8%AE%BEa%E4%B8%BA%E5%B8%B8%E6%95%B0%EF%BC%8C%E6%B1%82%E6%95%B0%E5%88%97a%EF%BC%8C2a2%EF%BC%8C3a2%EF%BC%8C%E2%80%A6%EF%BC%8Cnan%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%EF%BC%8E)
设数列a,2a2,3a2,…,nan的前n项和为Sn,
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−nan+1,
∴Sn=
a−an+1
(1−a)2−
nan+1
1−a,(a≠1)
若a=0,则Sn=0适合上式.
Sn=
n(n+1)
2,n=1
a−an+1
(1−a)2−
nan+1
1−a,n≠1.
∴数列a,2a2,3a2,…,nan的前n项和为
n(n+1)
2,n=1
a−an+1
(1−a)2−
nan+1
1−a,n≠1..
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−nan+1,
∴Sn=
a−an+1
(1−a)2−
nan+1
1−a,(a≠1)
若a=0,则Sn=0适合上式.
Sn=
n(n+1)
2,n=1
a−an+1
(1−a)2−
nan+1
1−a,n≠1.
∴数列a,2a2,3a2,…,nan的前n项和为
n(n+1)
2,n=1
a−an+1
(1−a)2−
nan+1
1−a,n≠1..
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