数列{an}的前n项和为Sn,已知Sn=(c+1)-can,常数c满足c(c+1)≠0.
(1) 求证{an}是等比数列
(2) 设数列{an}的公比q=f(C),数列{bn}满足b1=1/3,bn+1=f(bn)(n∈N*),试写出{1/bn}的通项公式,并求b1b2+b2b3+….+bn-1bn的值
(3) 设数列{1/bn}的前n项之和为Tn,是否存在非零常数λ使{(4Tn-11n)/n+λ}为等差数列,若存在,求出λ值;若不存在,请说明理由

数列{an}的前n项和为Sn,已知Sn=(c+1)-can,常数c满足c(c+1)≠0.
(1) 求证{an}是等比数列
(2) 设数列{an}的公比q=f(c),数列{bn}满足b1=1/3,b(n+1)=f(bn)(n∈N*),试写出{1/bn}的通项公式,并求b1b2+b2b3+…+bn-1bn的值
(3) 设数列{1/bn}的前n项之和为Tn,是否存在非零常数λ使{(4Tn-11n)/n+λ}为等差数列,若存在,求出λ值;若不存在,请说明理由

（1）Sn =(c+1)-can, S(n-1) =(c+1)-ca(n-1)
--->an = Sn-S(n-1) = ca(n-1)-can
--->(c+1)an = ca(n-1)
--->an/a(n-1) = c/(c+1) ......非零常数
--->{an}是等比数列

（2）q = f(x) = c/(c+1)
b(n+1)=f(bn)=bn/(bn+1)--->1/b(n+1)=1+1/bn
--->{1/bn}是等差数列，1/b1=3,d=1
--->1/bn = n+2--->bn=1/(n+2)
　b1b2+b2b3+...+b(n-1)bn
= 1/3•4 + 1/4•5 + 1/5•6 + ...+1/[(n+1)(n+2)]
= (1/3-1/4)+(1/4-1/5)+...+[1/(n+1)-1/(n+2)]
= 1/3-1/(n+2)
= (n-1)/(3n+6)

（3）--->Tn = 3+4+...+(n+2) = n(n+5)/2
--->Kn = (4Tn-11n)/n+λ = (2n-1)+λ = (1+λ)+2(n-1)
要使{Kn}是等差数列--->K(n+1)-Kn = 2为常数，显然成立
即λ可为任意实数