Question

已知等比数列{an}满足：a2=4，公比q=2，数列{bn}的前n项和为Sn，且Sn=4/3bn-2/3an+2/3

而n属于正整数 设cn=bn/an 求证c1/c2+c2/c3+…+cn/cn+1<n/2
an bn我会求只需说怎么证就可以了

Answer 1

a1=a2/q = 2
an =a1.q^(n-1) = 2^n

Sn = (4/3)bn -(2/3)an + 2/3
= (4/3)bn -(2/3).2^n + 2/3
n=1, b1= 2

bn = Sn -S(n-1)
=(4/3)bn - (4/3)b(n-1) - (1/3).2^n
bn = 4b(n-1) + 2^n
bn +2^n = 4[ b(n-1) +2^(n-1) ]
=>{bn +2^n} 是等比数列, q=4

bn +2^n = 4^(n-1) .[b1 +2^1]
=4^n
bn = 4^n-2^n

cn = bn/an
= (4^n -2^n)/2^n
= 2^n -1
cn /c(n+1) = (2^n -1)/(2^(n+1) -1) < 1/2
c1/c2+c2/c3+...+cn/c(n+1) < n/2

Answer 2

不懂可以追问,一定尽力解答,祝愉快

Answer 3

不懂可以追问,一定尽力解答,祝愉快

希望能解决您的问题。