已知数列{an}的前n项和为Sn,求{an}的通项公式.(1)Sn=2n²-3n+k (2)Sn=3²+b
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![已知数列{an}的前n项和为Sn,求{an}的通项公式.(1)Sn=2n²-3n+k (2)Sn=3²+b](/uploads/image/z/15095842-34-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%B1%82%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.%281%29Sn%3D2n%26sup2%3B-3n%2Bk+%282%29Sn%3D3%26sup2%3B%2Bb)
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已知数列{an}的前n项和为Sn,求{an}的通项公式.(1)Sn=2n²-3n+k (2)Sn=3²+b
已知数列{an}的前n项和为Sn,求{an}的通项公式.(1)Sn=2n²-3n+k (2)Sn=3²+b
已知数列{an}的前n项和为Sn,求{an}的通项公式.(1)Sn=2n²-3n+k (2)Sn=3²+b
a1=s1=2-3+k=k-1
Sn=2n^2-3n+k
S(n-1)=2(n-1)^2-3(n-1)+k
=2n^2-4n+2-3n+3+k
=2n^2-7n+5+k
an=sn-s(n-1)
=2n^2-3n+k-(2n^2-7n+5+k)
=2n^2-3n+k-2n^2+7n-5-k
=-4n-5
an=4n-5(n>=2)
a1=a1=9+b
Sn=3²+b=9+b
s(n-1)=9+b
an=sn-s(n-1)=0
an=0(n>=2)
(1)an=Sn-Sn-1=2n²-3n+k-2(n-1)²+3(n-1)-k=4n+5,与推论相符,当n=1时,a1=9推出k=10
∴an=4n+5
(2)an=Sn-Sn-1=0,∴数列an为恒为0的等差数列,即an=0,推出b=-9