Question

第16题怎么做，求详细过程

Answer 1

an=a1.q^(n-1) ; q<1
a2=3

a1+a3=10
a1.a2+a2.a3+....+an.a(n+1) = ?

solution:
a2=3
a1q=3 (1)

a1+a3=10
a1(1+q^2) =10 (2)
(2)/(1)
(1+q^2)/q = 10/3
3+3q^2 =10q
3q^2-10q+3 =0
(3q-1)(q-3)=0
q=1/3

from (1)
a1(1/3)=3
a1= 9

=>an = 9.(1/3)^(n-1)

cn
=an.a(n+1)
=9.(1/3)^(n-1) . 9.(1/3)^n
=27.(1/3)^(2n)

a1.a2+a2.a3+....+an.a(n+1)
=c1+c2+....+cn
=3[ 1 - (1/3)^2n ]/(1- 1/9)
=(27/8)[ 1 - (1/3)^2n ]