已知数列{an}满足a1=0,an+1+sn=n^2+2n已知数列{an}满足a1=0,a(n+1)+sn=n^2+2n(n属于N*),其中Sn为{an}的前n项和,求次数列的通项公式
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![已知数列{an}满足a1=0,an+1+sn=n^2+2n已知数列{an}满足a1=0,a(n+1)+sn=n^2+2n(n属于N*),其中Sn为{an}的前n项和,求次数列的通项公式](/uploads/image/z/3198775-31-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D0%2Can%2B1%2Bsn%3Dn%5E2%2B2n%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D0%2Ca%EF%BC%88n%2B1%EF%BC%89%2Bsn%3Dn%5E2%2B2n%28n%E5%B1%9E%E4%BA%8EN%2A%29%2C%E5%85%B6%E4%B8%ADSn%E4%B8%BA%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E6%B1%82%E6%AC%A1%E6%95%B0%E5%88%97%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
已知数列{an}满足a1=0,an+1+sn=n^2+2n已知数列{an}满足a1=0,a(n+1)+sn=n^2+2n(n属于N*),其中Sn为{an}的前n项和,求次数列的通项公式
已知数列{an}满足a1=0,an+1+sn=n^2+2n
已知数列{an}满足a1=0,a(n+1)+sn=n^2+2n(n属于N*),其中Sn为{an}的前n项和,求次数列的通项公式
已知数列{an}满足a1=0,an+1+sn=n^2+2n已知数列{an}满足a1=0,a(n+1)+sn=n^2+2n(n属于N*),其中Sn为{an}的前n项和,求次数列的通项公式
a(n+1)+Sn=S(n+1)=n^2+2n
S(n+1)+1=(n+1)^2
Sn=n^2-1 .(n>1)
S(n+1)-Sn=2n+1=a(n+1)
an=2n-1...(n>1)
an=0...(n=1)
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