Question

设数列{an}的前n项和为Sn,已知a1=1,a(n+1)=Sn((n+2)/n)(n=1,2,3,…).证明：

Answer 1

a(n+1)=Sn((n+2)/n)

s(n+1)-sn=Sn((n+2)/n)

s(n+1)=sn*2(n+1)/n

s(n+1)/(n+1)=2*sn/n

s1=a1=1,s1/1=1,

所以数列{Sn/n}是首项为1公比为2的等比数列.

2.s(n+1)/(n+1)=2^(n+1)-1,s(n+1)=(2^(n+1)-1)(n+1)

4an=4*s(n-1)*(n+1)/(n-1)=4(2^(n-1)-1)(n+1)