[好心人帮帮忙]设在等比数列{an}的前n项和为Sn,已知a(n+1)=2Sn+2（1）求数列{an}的通项公式

a(n)=aq^(n-1)
2a(1)+2=2a+2=2s(1)+2=a(2)=aq
2s(n)+2=a(n+1)=s(n+1)-s(n)
s(n+1)=3s(n)+2
s(n+1)+1=3[s(n)+1]
{s(n)+1}是首项为s(1)+1=a(1)+1=a+1,公比为3的等比数列.
s(n)+1=(a+1)*3^(n-1)
s(n+1)+1= (a+1)*3^(n)
a(n+1)=s(n+1)+1-s(n)-1=2(a+1)3^(n-1)=[2(a+1)/3]3^n
q=3
2a+2=aq=3a,a=2
a(n)=2*3^(n-1)
(2)
a(n+1)=2*3^(n)=a(n)+(n+1)d(n)=2*3^(n-1)+(n+1)d(n)
d(n)=[4/(n+1)]*3^(n-1)
1/d(n)=(1/4)(n+1)/3^(n-1)
t(n)=1/d(1)+1/d(2)+1/d(3)+...+1/d(n-1)+1/d(n)=(1/4)[2/1 + 3/3 + 4/3^2 + ...+ n/3^(n-2) + (n+1)/3^(n-1)]
t(n)/3=(1/4)[2/3 + 3/3^2 + 4/3^3 + ...+ n/3^(n-1) + (n+1)/3^n]
(2/3)t(n)=(1/4)[2/1+1/3+1/3^2+...+1/3^(n-1)-(n+1)/3^n]
=(1/4)[1+1/3+1/3^2+...+1/3^(n-1) + 1-(n+1)/3^n]
=(1/4)[(1-1/3^n)/(1-1/3) + 1 - (n+1)/3^n]
=(1/4)[(3/2)(1-1/3^n) + 1 - (n+1)/3^n]
t(n)=(3/8)[5/2 - (1/2)*3^(1-n) - (n+1)*3^(-n)]
(3)
d(n)=[4/(n+1)]*3^(n-1)
k=(m+p)/2
2(k-1)=(m+p-2)
(k+1)=[(m+1)+(p+1)]/2
[d(k)]^2=[4/(k+1)]^2*3^[2(k-1)]=d(m)*d(p)=[4/(m+1)][4/(p+1)]*3^(m+p-2)
(m+1)(p+1)=(k+1)^2=[(m+1)+(p+1)]^2/4
0=[(m+1)-(p+1)]^2
m=p=k
不存在这三项.