Question

高中数学题 数列

求 此题 第二问 谢谢要详细过程

Answer 1

如图所示

Answer 2

(1)
an =a1.q^(n-1)
Sn=a1+a2+...+an
a1=3
4S2 = 3S1 +S3
4a1.(1+q) = 3a1 + a1(1+q+q^2)
4(1+q) = 3 + (1+q+q^2)
q^2 -3q =0
q=3
an = 3^n
(2)
bn=log<3>an
= n
cn
= b(2n-1).b(2n) -b(2n).b(2n+1)
= (2n-1)(2n)-(2n)(2n+1)
=4n^2 -2n -4n^2 -2n
=-4n
b1.b2 -b2.b3 +b3.b4-b4.b5+.....+b(2n-1).b(2n)- b(2n).b(2n+1)
=c1+c2+...+cn
=-2n(n+1)

Answer 3

Answer 4

解答如上。