关于数列的数学问题

证明了Sn是数列{an}的前n项，A(1)= S(1)= 2A(1)-1 = >；A(1)=1

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所以B(n)是等比2的几何级数。

(2)b 1 = a 1-2n =-1所以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)

因此

tn=c1c2+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

当TN