关于数列 急

题目

关于数列 急
若等比数例满足：a1+a2+a3+a4+a5=3 a1²+a2²+a3²+a4²+a5²=12 则a1-a2+a3-a4+a5的值是多少?

答案

a1+a2+a3+a4+a5
=a1+q*a1+q^2*a1+q^3*a1+a^4*a1
=a1(1+q+q^2+q^3+q^4)
=a1(1-q^5)/(1-q)
=3——(1)
a1²+a2²+a3²+a4²+a5²
=a1²+q^2*a1²+q^4*a1²+q^6*a1²+q^8*a1²
=a1²(1+q^2+q^4+q^6+q^8)
=a1²(1-q^10)/(1-q^2)
=12——(2)
(2)/(1)
a1(1+q^5)/(1+q)=4
[1+q^5=(1+q)(1-q+q^2-q^3+q^4)]
a1(1-q+q^2-q^3+q^4)=4
a1-a2+a3-a4+a5
=a1-q*a1+q^2*a1-q^3*a1+a^4*a1
=a1(1-q+q^2-q^3+q^4)
=4