数列{an}是等比数列,则{kan}(k不等于0）,{1/an},{an^3},{an*a(n+1)},a(n+1)+an},{a(n+1)-an},{a(2n-1)},{n*an}中能构成等比数列的是?

设{an}公比为q;则
1
ka(n+1)/kan=a(n+1)/an=q;是等比数列;
2
[1/a(n+1)]/[1/an]=an/a(n+1)=1/q,是公比为1/q的等比数列;
3
a(n+1)^3/an^3=[a(n+1)/an]^3=q^3;是首相为a1^3,公比为q^3的等比数列;
4
[a(n+1)*a(n+2)]/[an*a(n+1)]=[a(n+1)/an]*[a(n+2)/a(n+1)]=q*q=q^2;
是首相为a1*a2,公比为q^2的等比数列;
5
[a(n+2)+a(n+1)]/[a(n+1)+an]=q*[a(n+1)+an]/[a(n+1)+an]=q.
是首相为a1+a2,公比为q的等比数列;
6
[a(n+2)-a(n+1)]/[a(n+1)-an]=q*[a(n+1)-an]/[a(n+1)-an]=q.
是首相为a2-a1,公比为q的等比数列;
7
a[2(n+1)-1]/a(2n-1)=a(2n+1)/a(2n-1)=q^2*a(2n-1)/a(2n-1)=q^2;
是首相为a1,公比为q^2的等比数列;
8
[(n+1)*a(n+1)]/(n*an)=[(n+1)*q*an]/(n*an)=q*(n+1)/n
q*(n+1)/n是随n变化的量,不是非零常数,所以{n*an}不是等比数列