Question

已知等差数列{an}满足a2=-7,a6+a8=6,求数列{an/2n-1}的前n项和?

Answer 1

已知等差数列{an}满足a2=-7,a6+a8=6,有

a2+4d+a2+6d=2a2+10d=6

得d=[6-2*(-7)]/10=2

得an=a2+(n-2)d=-7+2(n-2)=2n-11

所以an/2n-1=1-10/（2n-1)

数列{an/2n-1}的前n项和=n-10[1+1/3+1/5+...+1/(2n-1)]