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230问答网 > 在等比数列﹛an﹜中,a1+a2=1,a3+a4=2,则a5+a6+a7+a8=

在等比数列﹛an﹜中,a1+a2=1,a3+a4=2,则a5+a6+a7+a8=

2024-12-28 00:25:40

推荐回答（3个）

回答1：

a3+a4=2 --> q^2(a1+a2)=2, 又因为(a1+a2)=1，所以q^2=2;
所以a5+a6+a7+a8=q^4(a1+a2)+q^6(a1+a2)=1*4+1*8=12

回答2：

a2=a1q a3=a1q^2 a4=a1q^3

式1： a1+a2=a1(1+q)
式2： a3+a4=a1q^2(1+q) 式2除以式1 得q^=2 把q^2=2代入式1

最后就可以算后面的了。

回答3：

a1+a2=1,a3+a4=2

4d=a3+a4-(a1+a2)=1
a5+a6+a7+a8

=a3+a4+(a1+a2)+16d
=8

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