Question

已知等差数列{An}中,A(n+1)>An,A1A10=160,A3+A8=37

Answer 1

1.

等差数列,所以A1+A10=A3+A8=37

A1=5,A10=32

An=2+3n

2.

Sn=2*n+3*(2+4+8...+2^n)

=2n+3*(2^(n+1)-2)

=3*2^(n+1)+2n-6

3.

Gn=n(6+2+3n)=3n^2+8n

Sn-Gn=3*2^(n+1)-(6+6n+3n^2)

=3*{2^(n+1)-(2+2n+n^2)}

n=1,2,3,Sn=4,Sn>Gn

应该是高考题吧,后面求导我就省了