Question

已知数列an的通项公式为an=n^2,数列bn的首项b1=3,其前n项和为Sn,且满足关系式a(n+1)+a(n+2)+

Answer 1

a(n+1)+a(n+2)+...a(2n)

=(a1+a2+..a(2n))-(a1+a2+..an)

=(2n(2n+1)(4n+1)-n(n+1)(2n+1))/6

=(2n+1)(2n(4n+1)-n(n+1))/6

=(2n+1)( 7n^2+n)/6

Sn=n(2n+1)(7n+1)/6/((7n+1)/6)

=n(2n+1)

bn=Sn-S(n-1)=n(2n+1)-(n-1)(2n-1)=2n^2+n-(2n^2-3n+1)=

4n-1

q=2^(-bn)/2^(-b(n-1))=2^(1-4n-(1-4(n-1)))=2^(-4)=1/16

Tn=2^(-b1)*(1-q^n)/(1-q)>31/240

(1-q^n)/(1-q)>31/30

(1-(1/16)^n)>31/32

(1/16)^n1