求1,3a,5a2,7a3,…(2n-1)an-1的前n项和.
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求1,3a,5a2,7a3,…(2n-1)an-1的前n项和.
当a=1时,数列变为1,3,5,7,…,(2n-1),则Sn=
n[1+2(n−1)]
2=n2.
当a≠1时,有,
Sn=1+3a+5a2+7a3+…+(2n-1)an-1,①
aSn=a+3a2+5a3+7a4+…+(2n-1)an.②
①-②得Sn-aSn=1+2a+2a2+2a3+…+2an-1-(2n-1)an,
(1-a)Sn=1-(2n-1)an+2(a+a2+a3+a4+…+an-1)
=1-(2n-1)an+2•
a(1−an−1)
1−a
=1-(2n-1)an+
2(1−an)
1−a.
又1-a≠0,
∴Sn=
1−(2n−1)an
1−a+
2(a−an)
(1−a)2.
综上,Sn=
n2,a=1
1−(2n−1)an
1−a+
2(a−an)
(1−a)2,a≠1.
n[1+2(n−1)]
2=n2.
当a≠1时,有,
Sn=1+3a+5a2+7a3+…+(2n-1)an-1,①
aSn=a+3a2+5a3+7a4+…+(2n-1)an.②
①-②得Sn-aSn=1+2a+2a2+2a3+…+2an-1-(2n-1)an,
(1-a)Sn=1-(2n-1)an+2(a+a2+a3+a4+…+an-1)
=1-(2n-1)an+2•
a(1−an−1)
1−a
=1-(2n-1)an+
2(1−an)
1−a.
又1-a≠0,
∴Sn=
1−(2n−1)an
1−a+
2(a−an)
(1−a)2.
综上,Sn=
n2,a=1
1−(2n−1)an
1−a+
2(a−an)
(1−a)2,a≠1.
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