记数列(an)的前n项和为Sn已知a1=1,对任意n∈N*,均满足an+1=(n+2)/n)Sn求证:数列(Sn/n)为等比数列求数列(an)的通项公式
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![记数列(an)的前n项和为Sn已知a1=1,对任意n∈N*,均满足an+1=(n+2)/n)Sn求证:数列(Sn/n)为等比数列求数列(an)的通项公式](/uploads/image/z/8618270-14-0.jpg?t=%E8%AE%B0%E6%95%B0%E5%88%97%28an%29%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%E5%B7%B2%E7%9F%A5a1%3D1%2C%E5%AF%B9%E4%BB%BB%E6%84%8Fn%E2%88%88N%2A%2C%E5%9D%87%E6%BB%A1%E8%B6%B3an%2B1%3D%28n%2B2%29%2Fn%29Sn%E6%B1%82%E8%AF%81%EF%BC%9A%E6%95%B0%E5%88%97%EF%BC%88Sn%2Fn%29%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%E6%B1%82%E6%95%B0%E5%88%97%EF%BC%88an%EF%BC%89%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
记数列(an)的前n项和为Sn已知a1=1,对任意n∈N*,均满足an+1=(n+2)/n)Sn求证:数列(Sn/n)为等比数列求数列(an)的通项公式
记数列(an)的前n项和为Sn已知a1=1,对任意n∈N*,均满足an+1=(n+2)/n)Sn
求证:数列(Sn/n)为等比数列
求数列(an)的通项公式
记数列(an)的前n项和为Sn已知a1=1,对任意n∈N*,均满足an+1=(n+2)/n)Sn求证:数列(Sn/n)为等比数列求数列(an)的通项公式
证明,
因为A(n+1) = (n+2)/n * Sn
所以Sn = n*A(n+1) / (n+2)
S(n-1) = (n-1)*An / (n+1)
所以An = Sn - S(n-1) = n/(n+2) *A(n+1) - (n-1)/(n+1) * An
所以2n/(n+1) * An = n/(n+2) * A(n+1)
即A(n+1)/An = (2n+4)/(n+1)
所以(Sn/n) / (S(n-1)/(n-1)) = ( A(n+1)/(n+2) ) / ( An / (n+1))
= A(n+1)/An * (n+1)/(n+2)
= (2n+4)/(n+1) * (n+1)/(n+2) = 2
所以Sn/n是以2为公比的等比数列
(2)
因为Sn/n是以2为公比的等比数列,首项为S1/1=S1=A1=1
所以Sn/n的通项公式是2^(n-1)
所以Sn = n*2^(n-1)
S(n-1) = (n-1)*2^(n-2)
所以An = Sn - S(n-1) = n*2^(n-1) - (n-1)*2^(n-2)
= n*2^(n-1) - n*2^(n-2) + 2^(n-2)
= n*2^(n-2) + 2^(n-2)
= (n+1) * 2^(n-2)
当n=1时也满足,所以通项公式为An = (n+1) * 2^(n-2)