已知数列{an}满足a1=0,an+1 +Sn=n2+2n(n属于N*),其中Sn为{an}的前n项的和,求此数列的通项
已知数列{an}满足a1=0,an+1 +Sn=n2+2n(n属于N*),其中Sn为{an}的前n项的和,求此数列的通项公式.
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a(n+1)+Sn=n^2+2n
an+S(n-1)=(n-1)^2+2(n-1)
两式相减得
a(n+1)+Sn-an-S(n-1)=n^2+2n-(n-1)^2-2(n-1)
a(n+1)+an-an=n^2+2n-n^2+2n-1-2n+2
a(n+1)=2n+1
a(n+1)=2(n+1)-1
an=2n-1
a(n+1) +Sn=n^2+2n
Sn=n^2+2n-a(n+1)
=n^2+2n-(2n+1)
=n^2+2n-2n-1
=n^2-1
an=sn-s(n-1)
=n^2-1-[(n-1)^2-1]
=n^2-1-n^2+2n
=2n-1
an+S(n-1)=(n-1)^2+2(n-1)
两式相减得
a(n+1)+Sn-an-S(n-1)=n^2+2n-(n-1)^2-2(n-1)
a(n+1)+an-an=n^2+2n-n^2+2n-1-2n+2
a(n+1)=2n+1
a(n+1)=2(n+1)-1
an=2n-1
a(n+1) +Sn=n^2+2n
Sn=n^2+2n-a(n+1)
=n^2+2n-(2n+1)
=n^2+2n-2n-1
=n^2-1
an=sn-s(n-1)
=n^2-1-[(n-1)^2-1]
=n^2-1-n^2+2n
=2n-1
1年前
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