问题详情:
数列{an}满足a1=,a2=,且a1a2+a2a3+…+anan+1=na1 an+1对任何正整数n成立,则
++…+的值为 .
【回答】
85 解法一:由a1a2+a2a3=2a1a3及a1=,a2=,得a3=,再由a1a2+a2a3+ a3a4=3a1a4,a4=.
进一步得a5=,a6=, a7=,a8=,a9=,a10=,故++…+=4+5+6+7+8+9+10+11+12+13=85.解法二:由a1a2+a2a3+…+anan+1=na1 an+1 ①,a1a2+a2a3+…+anan+1+ an+1an+2=(n+1)a1 an+2 ②,②-①得,an+1an+2=(n+1)a1 an+2-na1 an+1=-=-=+,(n≥2),则a1a2+a2a3=2a1a3=+,所以数列{}成等差数列,公差为1,即=n+3,an=.代入可得++…+=85.
知识点:数列
题型:填空题