已知数列{an},a1=1,a3=4,其前n项和Sn满足Sn+1=2入Sn十1(入是大于0的常数),n属于N*1.求入的值并证明{Sn十1}是等比数列2.设数列{nan}的前n项和为Tn,试比较Tn/2与Sn的大小

Question

已知数列{an},a1=1,a3=4,其前n项和Sn满足Sn+1=2入Sn十1(入是大于0的常数),n属于N*
1.求入的值并证明{Sn十1}是等比数列
2.设数列{nan}的前n项和为Tn,试比较Tn/2与Sn的大小

Answer 1

(1)S(n+1) =2入Sn +1S(n+1) + 1 = 2入(Sn +1)coef. of constant(2入-1) = 1入= 1(2)入= 1S(n+1) + 1 = 2(Sn +1)[S(n+1) + 1]/(Sn +1) =2(Sn +1)/(S1+1) = 2^(n-1)Sn = 2^n -1 (1)S(n-1) = 2^(n-1) ...