Question

等比数列解方程组该怎么解？

这个方程组怎么解出来，求详细过程

Answer 1

你问得应该是第一问吧，这个不需要解出具体数值，解出Cn关于An的表达式形式，比值为定值即可证明，

计算过程如图，你写过解题的部分就不写了

Answer 2

an=a1.q^(n-1), Sn=a1+a2+...+an

(1)
an+Sn = n (1)
a(n-1)+S(n-1) = n-1 (2)
(1)-(2)
an -a(n-1) + an = 1
an = (1/2)a(n-1) +1/2
an -1 = (1/2)[ a(n-1) -1 ]
=> cn = an -1 是等比数列, q=1/2
(2)
an -1 = (1/2)^(n-1) . ( a1 -1 )
an = 1+ (1/2)^(n-1) . ( a1 -1 )
bn
=an - a(n-1)
=1+ (1/2)^(n-1) . ( a1 -1 ) - [ 1+ (1/2)^(n-2) . ( a1 -1 ) ]
=-(a1-1)(1/2)^(n-2)
b1 = a1
a1 =-2(a1-1)
a1= 2/3
ie
bn = (1/3).(1/2)^(n-2)