已知数列{an}的前n项的和Sn=n^2+5n-1求通项公式在等差数列{an}中,已知d=2,an=11 Sn=35,求an和n

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已知数列{an}的前n项的和Sn=n^2+5n-1求通项公式
在等差数列{an}中,已知d=2,an=11 Sn=35,求an和n

a1=s1=1^2+5-1=5
sn=n^2+5n-1
s(n-1)=(n-1)^2+5(n-1)-1
=n^2-2n+1+5n-5-1
=n^2+3n-5
an=sn-s(n-1)
=n^2+5n-1-(n^2+3n-5)
=n^2+5n-1-n^2-3n+5
=2n+4
an=2n+4(n>=2)
an=a1+(n-1)d
11=a1+2(n-1)
a1+2n-2=11
a1=13-2n
sn=(a1+an)*n/2
(a1+11)*n/2=35
(a1+11)*n=70
(13-2n+11)*n=70
(24-2n)*n=70
2n^2-24n+70=0
n^2-12+35=0
(n-5)(n-7)=0
n=5或n=7
当n=5时
a1=13-2n=13-2*5=3
当n=7时
a1=13-2n=13-2*7=-1

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