等差数列的前n项和已知等比数列{an}中,a2=2,a5=1/4,求和：a1a2+a2a3+…+anan+1.

a5=a2q^31/4=2q^3q^3=1/8q=1/2a2=a1q2=a1*1/2a1=4an=a1q^(n-1)=4*(1/2)^(n-1)=(1/2)^(n-3)an*a(n+1)=(1/2)^(n-3)*(1/2)^(n-2)an*a(n+1)=(1/2)^(2n-5)a1a2+a2a3+…+ana(n+1)=(1/2)^(-3)+(1/2)^(-1)+(1/2)^1+。+(1/2)^(2n-5)=8*[1-(1/2^2)^n]/(1-1/2^2)=8*[1-(1/2)^2n]/(1-1/4)=8*[1-(1/2)^2n]/(3/4)=32*[1-(1/2)^2n]/3=[32-(1/2)^(2n-5)]/3