高二数学题

已知数列{an}的前n项和为Sn,且a1=1,an+1=Sn(n+2)/n ,(n 属于 N*)求(1)数列{Sn/n}是等比数列(2)Sn+1=4an 首先从 数列本身的基本意义出发a = S - S 其次 ，从已知a=S(n+2)/n 出发a= S * (n+1)/(n-1)因此S - S = S * (n+1)/(n-1)移项整困信仔理S = S/(n-1)因此 S/n 是等比数列。公比为2。=====================S<1>/1 = a1/1 = 1因此 S/n = (S<1>/1)*q^(n-1) = 1 * 2^(n-1) = 2^(n-1)S = n* 2^(n-1)S = (n+1) * 2^n因为 a = S - S所以 a = (n+1)*2^n - n*2^(n-1)将下角标的 n+1替换为n, 将n替换为 (n-1),得到a = n*2^(n-1) - (n-1) * 2^(n-2)4a = n*2^(n+1) - (n-1) * 2^n= 2n * 2^n - (n-1) * 2^n= (n+1) * 2^n = S