如题~设数列前n项和sn……a1+2a2+3a3+……+nan=(n―1)sn+2n……求an通项公式
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如题~设数列前n项和sn……a1+2a2+3a3+……+nan=(n―1)sn+2n……求an通项公式
![如题~设数列前n项和sn……a1+2a2+3a3+……+nan=(n―1)sn+2n……求an通项公式](/uploads/image/z/7929899-35-9.jpg?t=%E5%A6%82%E9%A2%98%7E%E8%AE%BE%E6%95%B0%E5%88%97%E5%89%8Dn%E9%A1%B9%E5%92%8Csn%E2%80%A6%E2%80%A6a1%2B2a2%2B3a3%2B%E2%80%A6%E2%80%A6%2Bnan%3D%28n%E2%80%951%29sn%2B2n%E2%80%A6%E2%80%A6%E6%B1%82an%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
n用n+1代,则a1+……(n+1)an+1=nSn+1+2(n+1)
两式相减得(n+1)an+1=n(Sn+1-Sn)+Sn+2
而Sn+1-Sn=an+1,所以an+1=Sn +2.
n用n-1代,得an=Sn-1 +2;
两式相减得an+1-an=an所以an+1=2an
n用1代a1=2.所以an=2^(n-1)a1=2^n
PS:打的怎么辛苦,要采纳啊
两式相减得(n+1)an+1=n(Sn+1-Sn)+Sn+2
而Sn+1-Sn=an+1,所以an+1=Sn +2.
n用n-1代,得an=Sn-1 +2;
两式相减得an+1-an=an所以an+1=2an
n用1代a1=2.所以an=2^(n-1)a1=2^n
PS:打的怎么辛苦,要采纳啊
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