An=S(n)-S(n- 1)
=2a(n)+n^2-3n-2-[2a(n- 1)+(n- 1)^2-3(n- 1)-2]
= 2(A(n)-A(n- 1)+2n-4
= = & gtA(n)=2A(n- 1)-2n+4
= = & gtA(n)-2n=2A(n- 1)-4n+4
=2[A(n- 1)-2(n- 1)]
= = & gtB(n)=2B(n- 1)
S So B(n) is a geometric series with an equal ratio of 2.
(2) b1= a1-2n =-1so b (n) =-2 (n- 1).
c(n)= 1/b(n+ 1)=- 1/2^n
= = & gtc(n)c(n+ 1)=- 1/2^(n)* - 1/2^(n+ 1)
= 1/2^(2n+ 1)
= = & gt2^(n- 1)c(n)c(n- 1)= 1/2^(n+2)
therefore
tn=c 1c2+2c2c3+2^2c3c4+…+2^(n- 1)cnc(n+ 1)
= 1/2^3+ 1/2^4 +…………+ 1/2^(n+2)
= 1/2^3[ 1+ 1/2+ 1/2^2+…………+ 1/2^(n- 1)]
= 1/8 *[( 1- 1/2^n)/( 1- 1/2)]
= ( 1- 1/2^n)/4
When TN