设数列{bn}的前n项和为sn,且bn=2-2sn;数列{an}为等差数列,且a5=14,a7=20求

1 = 2-2*b1
3b1 = 2
b1 = 2/3
bn - bn-1 = (2-2sn) - (2-2sn-1) = -2(sn- sn-1) = -2bn
3bn = bn-1
bn = 1/3 * bn-1
{bn}是等比数列
{bn} = { 2/3 * (1/3)^(n-1) } = { 2 * (1/3)^n }
a7 - a5 = 2d
d = (20 - 14 )/2 = 3
a1 = a5 - 4d = 14 - 4*3 = 2
{an} = { 2+ 3(n-1)} = { 3n -1 }
cn = an * bn = (2* (1/3)^n )*(3n-1)) = (6n-2) (1/3)^n
Tn = c1 + c2 +c3 + ...+ cn
Tn - 1/3 Tn =( c1 + c2 +c3 + ...+ cn) - 1/3(c1 + c2 +c3 + ...+ cn)
= c1 + (c2 - 1/3 c1) + (c3- 1/3 c2) + ...+ (cn -1/3 cn-1) - cn
= 4/3 + 6 * (1/3)^2 + 6 *(1/3)^3 + ...+ 6 * (1/3)^n - (6n-2)*(1/3)^n
= 4/3 + 6*( (1/3)^2 - (1/3)^(n+1)) / (1-1/3) ) -(6n-2)*(1/3)^n
= 4/3 + ( 1- (1/3)^(n-1) ) - (6n-2)*(1/3)^n
= 7/3 - (1/3)^(n-1) - (6n-2)*(1/3)^n
= 7/3 - 3*(1/3)^n -6n *(1/3)^n + 2 *(1/3)^n
= 7/3 - (1/3)^n - 6n *(1/3)^n
< 7/3
所以 2/3 *Tn < 7/3
Tn < 7/3 * 3/2 = 7/2
再问： Tn - 1/3 Tn =( c1 + c2 +c3 + ... + cn) - 1/3(c1 + c2 +c3 + ... + cn) （1） = c1 + (c2 - 1/3 c1) + (c3- 1/3 c2) + ... + (cn -1/3 cn-1) - cn （2） 第（2）步骤我得是：=c1 + (c2 - 1/3 c1) + (c3- 1/3 c2) + ... + (cn -1/3 cn-1) -1/3cn
再答： 你是对的， 我这里漏了系数1/3 应当是 Tn - 1/3 Tn =( c1 + c2 +c3 + ... + cn) - 1/3(c1 + c2 +c3 + ... + cn) = c1 + (c2 - 1/3 c1) + (c3- 1/3 c2) + ... + (cn -1/3 cn-1) - 1/3 cn = 4/3 + 6 * (1/3)^2 + 6 *(1/3)^3 + ... + 6 * (1/3)^n - (1/3) (6n-2)*(1/3)^n = 4/3 + 6*( (1/3)^2 - (1/3)^(n+1)) / (1-1/3) ) -(6n-2)*(1/3)^(n+1) = 4/3 + ( 1- (1/3)^(n-1) ) - (6n-2)*(1/3)^(n+1) = 7/3 - (1/3)^(n-1) - (6n-2)*(1/3)^(n+1) = 7/3 - 9*(1/3)^(n+1) -6n *(1/3)^n + 2 *(1/3)^(n+1) = 7/3 - 7 *(1/3)^n - 6n *(1/3)^n < 7/3