Question

在数列{an}中,a1＝3,且对于任意大于1的正整数n,点(an,an－1)在直线x-y=√6上.(1),求数列a7的通

Answer 1

(1) an-an-1=6 a1=3 an是以3为首项6为公差的等差数列,an=3+(n-1)*6=6n-3a7=6*7-3=39(2) 设bn=a7/3n(n+1)=13/n(n+1)=13[1/n-1/(n+1)]sn=b1+b2+b3+...+bn=13[1/1-1/2+1/2-3-1/3+...+1/n-1/(n-1)]=13(1-1/n)...