题目

{an}为等差数列,且Sn/Sk=n^2/k^2,则an/ak=

提问时间：2020-07-26

答案

设an=a1+(n-1)a Sn=na1+n(n-1)a/2 Sk=ka1+k(k-1)a/2 Sn/Sk=[na1+n(n-1)a/2]/[ka1+k(k-1)a/2]=n^2/k^2 kn^2a1+k^2n(n-1)a/2=kn^2a1+k(k-1)n^2a/2 2kn^2a1-k^2na=2kn^2a1-kn^2a 2kna1(n-k)-kna(k-n)=0 n=k或2kna1+kna=0 2a1+a=0 a1/a=-1/2 an/ak=1（n=k时） an/ak=（a1+(n-1)a）/（a1+(k-1)a=(-1/2+n-1)/(-1/2+k-1)=(2n-3)/(2k-3)