Question

求解第二小题的通项公式！速求答案！麻烦各位大神.学霸了

Answer 1

Sn=Tn-Tn-1=2Sn-n²-(2Sn-1-(n-1)²)
=2Sn-2Sn-1-2n+1=2(Sn-Sn-1)-2n+1
An=Sn-Sn-1
Sn=2An-2n+1
An=Sn-Sn-1=2An-2n+1-(2An-1-2(n-1)+1)=2(An-An-1)-2
An=2An-1+2
An+2=2(An-1+2)
令Cn=An+2
Cn=2Cn-1,C1=A1+2=3
Cn/Cn-1=2
Cn=3*2^(n-1)
An=3*2^(n-1)-2

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Answer 2

Sn=a1+a2+……+an
Tn=S1+S2+……+Sn

Tn-T(n-1)=2(Sn-S(n-1))-n^2+(n-1)^2
Sn=2an-2n+1 S(n-1)=2a(n-1)-2(n-1)+1
Sn-S(n-1)
=2an-2n+1-[2a(n-1)-2(n-1)+1]
=2an-2a(n-1)-2
an=2an-2a(n-1)-2

an=2a(n-1)+2
an+2=2[a(n-1)+2]
数列{an+2}是等比数列 首项=a1+2=3，公比=2
an+2=3*2^(n-1)

an=3*2^(n-1)-2