Question

一道数学数列,函数题已知各项均不为0的数列｛an｝的前k项和为Sk,且Sk=ak ×ak+1／2（ak和ak+1是第k项

Answer 1

Sk=ak×a(k+1)/2

Sk-1=ak×a(k-1)/2

ak=Sk-Sk-1=ak[a(k+1)-a(k-1)]/2

a1=1≠0,又ak=ak[a(k+1)-a(k-1)]/2对于任意正整数k均成立,因此ak≠0

等式两边同除以ak

[a(k+1)-a(k-1)]/2=1

a(k+1)=a(k-1)+2

a3=a1+2=1+2=3

S2=a1+a2=(a2a3)/2

1+a2=3a2/2 a2=2

an=n

数列{an}的通项公式为an=n.

(an)^(-a)