Question

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

Answer 1

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)

谢谢采纳 ^_^