Question

记数列(an)的前n项和为Sn已知a1=1,对任意n∈N*,均满足a（n+1）=(n+2)/n)Sn求证：数列（Sn/n)为等比数列 求

Answer 1

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)

Answer 2

A(n+1)=S(n+1)-Sn 所以S(n+1)-Sn=[(n+2)/n]Sn 将Sn移到右边可得S(n+1)=[(n+1)/n]Sn 两边同除n+1可得S(n+1)/(n+1)=2Sn/n,所以Sn/n是等比数列
由a（n+1）=(n+2)/n)Sn可知数列Sn/n是S2/2以为首项，令n=1代入上式，解得S2=4,所以Sn/n是以2为首项,2为公比的等比数列
所以Sn=n2^(n-1),A(n+1)=S(n+1)-Sn =(n+1)2^n-n2^(n-1)=(n+2)2^(n-1),
所以An=(n+1)2^(n-2),n>1,
将A1也符合，所以An=(n+1)2^(n-2)