已知在等比数列{an}中,a1=2,q=1/2,则a1^2+a2^2+a3^2+.+an^2等于多少

Question

已知在等比数列{an}中,a1=2,q=1/2,则a1^2+a2^2+a3^2+.+an^2等于多少

Answer 1

an=a1*q^(n-1)
=2*(1/2)^(n-1)
=(1/2)^(n-2)
因此{（an)^2}是首项为1,公比为1/4的等比数列
an^2=[(1/2)^(n-2)]^2
=(1/4)^(n-2)
a1^2+a2^2+……+an^2
=[1-(1/4)^(n-1)]/(1-1/4)
=[1-(1/4)^(n-1)]/(3/4)
=[1-(1/2)^(2n-2)]/(3/4)
=4/3[1-(1/2)^(2n-2)]
=4/3-4/3(1/2)^(2n-2)
=4/3-4/3*2^(2-2n)
=4/3-2^(4-2n)/3