Question

已知{an}是等比数列,S2=2,S4=10,则a1a3+a2a4+…+a(n-1)a(n+1)+ana(n+2)=

Answer 1

a1+a2=2 a1+a2+a3+a4=10

a1(1+q)=2 a1(1+q+q^2+q^3)=10

a1=2/3 q=2 or a1=-2 q=-2

ana(n+2)=a1^2q^2n

a1a3+a2a4+…+a(n-1)a(n+1)+ana(n+2)

=a1^2q^2+a1^2q^4+...+a1^2q^2(n-1)+a1^2q^2n

=a1^2(q^2+q^4+.+q^2(n-1)+q^2n) q^2+q^4+.+q^2(n-1)+q^2n看成等比数列

=a1^2[q^2(1-q^2n)/(1-q^2)]

=16/27(2^2n-1) or16/3(2^2n-1)