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已知数列{an} {bn} {cn}满足(an+1-an)(bn+1-bn)=cn,n属于N*

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已知数列{an} {bn} {cn}满足(an+1-an)(bn+1-bn)=cn,n属于N*
(1)设an=1/3^n,bn=1-3n,求数列{cn}的前n项和Sn
(2)设cn=2n+4,{an}是公差为2的等差数列,若b1=1,求{bn}的通项公式
(3)设cn=3n-25,an=n^2-8n,求正整数k使得对一切n属于N*,均有bn≥bk
已知数列{an} {bn} {cn}满足(an+1-an)(bn+1-bn)=cn,n属于N*
c(n)=[a(n+1)-a(n)][b(n+1)-b(n)],
(1) c(n) = -3[1/3^(n+1)-1/3^n] = -3*1/3^(n+1)*[1-3] = 2/3^n,
s(n) = (2/3)[1+1/3 + ...+ 1/3^(n-1)] = (2/3)[1-1/3^n]/[1-1/3] = 1-1/3^n
(2) 2n+4 = 2[b(n+1)-b(n)],
b(n+1)-b(n) = n+2,
b(n+1) = b(n) + n+2 = b(n) + [n(n+1)-(n-1)n]/2 + 2[n+1-n],
b(n+1) - n(n+1)/2 - 2(n+1) = b(n) - (n-1)n/2 - 2n,
{b(n)-(n-1)n/2 - 2n}是首项为b(1)-2=-1,的常数数列.
b(n) - (n-1)n/2 -2n = -1,
b(n) = (n-1)n/2 + 2n-1
(3) 3n-25 = [(n+1)^2-n^2-8][b(n+1)-b(n)]=[2n-7-n^2][b(n+1)-b(n)],
n^2 -2n + 7 = (n-1)^2 + 6 >=6 >0.
b(n+1)-b(n) = (3n-25)/[2n-7-n^2] = 3(25/3-n)/[(n-1)^2 + 6],
1b(n),{b(n)}单调递增.1