Question

已知数列｛an｝的前n项和Sn满足Sn=3/2an+n-3。求证｛an-1｝是等比数列

Answer 1

Sn=3/2an+n-3
an=Sn-Sn-1
an=3/2(an-an-1)+1
an=3an-1-2
an-1=3(an-1-1)
(an-1)/(an-1-1)=3
A1=a1-1
S1=a1-1=4-1=3
｛an-1｝是等比数列,公比=3,首项=4
an-1=3^n
an=(3^n)+1

Answer 2

a(1)=s(1)=3a(1)/2 + 1-3 ,
a(1) = 4.

s(n) = 3a(n)/2 + n-3,
s(n+1) = 3a(n+1)/2 + (n+1)-3,
a(n+1) = s(n+1)-s(n) = 3a(n+1)/2 - 3a(n)/2 + 1,
a(n+1) = 3a(n) - 2,
a(n+1) -1 = 3a(n) - 3 = 3[a(n)-1],
{a(n)-1}是首项为a(1)-1=3,公比为3的等比数列。

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a(n) - 1 = 3*3^(n-1) = 3^n,
a(n) = 1 + 3^n

Answer 3

你这个不对吧，首项应该是3

Answer 4

不知道
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