问题
选择题
已知数列an=1+
|
答案
∵ak=1+
+1 2
+…+1 3
,ak+1=1+1 k2
+1 2
+…+1 3
+1 k2
+…+1 k2+1
,1 (k+1)2
∴ak+1-ak=
+…+1 k2+1
=1 (k+1)2
+1 k2+1
+…+1 k2+2
,1 k2+2k+1
∴共有k2+2k+1-(k2+1)+1=2k+1项.
故选D.
已知数列an=1+
|
∵ak=1+
+1 2
+…+1 3
,ak+1=1+1 k2
+1 2
+…+1 3
+1 k2
+…+1 k2+1
,1 (k+1)2
∴ak+1-ak=
+…+1 k2+1
=1 (k+1)2
+1 k2+1
+…+1 k2+2
,1 k2+2k+1
∴共有k2+2k+1-(k2+1)+1=2k+1项.
故选D.