2) because the point (bn, Sn) is on the straight line x+2y-2=0, there is bn+2Sn-2=0, so Sn=- 1/2*bn+ 1,
So s (n-1) =-1/2 * b (n-1)+1,the two expressions are subtracted, so.
Sn-s (n-1) =-1/2 (bn-b (n-1)), that is.
Bn=- 1/2(bn-b(n- 1)), available.
Bn= 1/3*b(n- 1), so the sequence {bn} is a geometric series.
And from b 1+2S 1-2=0, we can get that b 1+2b 1-2=0, b 1=2/3, and the general formula of sequence {bn} is
bn=b 1*( 1/3)^(n- 1)=2/3^n。
3) From the meaning of the question, cn = an * bn = (n+ 1) * 2/3 n, so there is.
TN = 2 * 2/31+2 * 3/32+...+2 * n/3 (n-1)+2 * (n+1)/3n, therefore.
1/3*Tn = 2*2/3^2+2*3/3^3+...+2*n/3^n+2*(n+ 1)/3^(n+ 1)，
Two kinds of dislocation subtraction, get.
2/3 * TN = 2*2/3^ 1+2* 1/3^2+...+2* 1/3^n-2*(n+ 1)/3^(n+ 1)
=2* 1/3^ 1+2( 1/3+ 1/3^2+...+ 1/3^n)-2*(n+ 1)/3^(n+ 1)
=2/3+2( 1/3- 1/3^(n+ 1))/( 1- 1/3)-2*(n+ 1)/3^(n+ 1)
=2/3+3( 1/3- 1/3^(n+ 1))-2*(n+ 1)/3^(n+ 1)
=2/3+ 1- 1/3^n-2*(n+ 1)/3^(n+ 1)
=5/3-(2n+5)/3^(n+ 1)
Therefore TN = (5/3-(2n+5)/3 (n+1))/(2/3).
=5/2-(n+5/2)/3^n
It is obvious that Tn≤5/2, and c 1=4/3, and any cn = an * bn>0, so Tn≥T 1=4/3, so 4/3≤Tn≤5/2.