Question

已知a,b,a+b成等差数列,a,b,ab成等比数列,则通项为a>n=4/2an^2+bn的数列{an}的前n项和为?

Answer 1

a,b,a+b成等差数列,2b=a+a+b b=2a

a,b,ab成等比数列 b^2=a*ab b=a^2

解得a=2 b=4

所以an=4/(2*2n^2+4n)=1/n(n+1)=1/n-1/(n+1)

{an}的前n项和=(1-1/2)+(1/2-1/3)+...+[1/n-1/(n+1)]

=1-1/(n+1)

=n/(n+1)