wayneqian
举报
整理得到:an-a(n-1)=(an-a1)/(n-1) ......[1] 同样有:a(n-1)-a(n-2)=[a(n-1)-a1]/(n-2) .....[2} [1]-[2]: an - 2a(n-1)+a(n-2) = (an-a1)/(n-1)- [a(n-1)-a1]/(n-2) = [nan-2an- na1+2a1-na(n-1)+a(n-1)+na1-a1]/(n-1)(n-2) "= {(n-1)[an-a(n-1)] - (an-a1)}/(n-1)(n-2) =[an-a1 - (an-a1)]/(n-1)(n-2)"=0 那么打引号的两步是怎么推证到的?