Question

等比数列{an}中,(a1+a3+a5)-(a2+a4)=4，(a1^2+a3^2+a5^2)+(a2^2+a4^2)=36

则（a1+a3+a5）+（a2+a4）＝？
过程也要啊...！

Answer 1

,(a1+a3+a5)-(a2+a4)=a3(q^2+1/q^2-q-1/q)=4
，(a1^2+a3^2+a5^2)+(a2^2+a4^2)=a3^2(q^4+1/q^4-q^2-1/q^2)=36
（a1+a3+a5）+（a2+a4）＝a3(q^2+1/q^2+q+1/q)=x
4x=a3^2[(q^2+1/q^2)+(q+1/q)][(q^2+1/q^2）-（q+1/q)]=a3^2(q^4+1/q^4-q^2-1/q^2)=36
x=9

Answer 2

(a1+a3+a5)-(a2+a4)
=(a1+a1p^2+a1p^4)-(a1p+a1p^3)
=a1(1-p+p^2-p^3+p^4)
=a1(p^5+1)/(p+1)=4
(a1^2+a3^2+a5^2)+(a2^2+a4^2)
=(a1^2+a1^2p^4+a1^2p^8)+(a1^2p^2+a1^2p^6)
=a1^2(1+p^2+p^4+p^6+p^8)
=a1^2(p^10-1)/(p^2-1)
=a1^2(p^5+1)(p^5-1)/(p+1)(p-1)
=[a1(p^5+1)/(p+1)][a1(p^5-1)/(p-1)]
=4a1(p^5-1)/(p-1)=36
∴a1(p^5-1)/(p-1)=9
a1(p^5-1)/(p-1)
=a1(1+p+p^2+p^3+p^4)
=(a1+a1p^2+a1p^4)+(a1p+a1p^3)
=(a1+a3+a5)+(a2+a4)
=9

Answer 3

你交换机你，不一 不会