问题
选择题
数列{an}满足:a1=
|
答案
a1a2+a2a3+…+anan+1=na1an+1,①
a1a2+a2a3+…+anan+1+an+1an+2=(n+1)a1an+2,②
①-②,得-an+1an+2=na1an+1-(n+1)a1an+2,
∴
-n+1 an+1
=4,n an+2
同理,得
-n an
=4,n-1 an+1
∴
-n+1 an+1
=n an+2
-n an
,n-1 an+1
整理,得
=2 an+1
+1 an
,1 an+2
∴{
}是等差数列.1 an
∵a1=
,a2=1 4
,1 5
∴等差数列{
}的首项是1 an
=4,公差d=1 a1
-1 a2
=5-4=1,1 a1
=4+(n-1)×1=n+3.1 an
∴
+1 a1
+…+1 a2
=97× 4+1 a97
×1=5044.97×96 2
故选B.