Question

已知各项全不为零的数列an的前K项和为Sk,且Sk＝1/2(akak+1(k ∈N*),其中a1＝1.

Answer 1

(1)s[k+1]-sk得

a[k+2]-a[k]=2

a[1]=1

a[2]=2

a[n]=n

(2)(1/n)^a=0