题目

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.

提问时间：2021-02-19

答案

a5=a2q^3
1/4=2q^3
q^3=1/8
q=1/2
a2=a1q
2=a1*1/2
a1=4
an=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