Question

已知｛an｝是等比数列，a2=2，a5=1/4,则a1a2+a2a3+......+anan+1=

急速度

Answer 1

解
设公比是q
a5=a2q^3=2q^3=1/4
解得q=1/2,a1=a2/q=4
ana(n+1)=a1q^(n-1)*a1q^n=(a1)²q^(2n-1)
a1a2+a2a3+......+anan+1
=(a1)²(q^1+q^3+...+q^(2n-1))
=16q(1-q^(2n))/(1-q^2)
=8(1-1/2^(2n))/(3/4)
=32(1-1/2^(2n))/3 (n>=1)