limn→∞[11•4+14•7+17•10+…+1(3n−2)(3n+1)]=______．
问题描述：
[lim n→∞
+1 1•4
+1 4•7
+…+1 7•10
]=______． 1 (3n−2)(3n+1)
答
[lim n→∞
+1 1•4
+1 4•7
+…+1 7•10
]1 (3n−2)(3n+1)
=
lim n→∞
[(1−1 3
)+(1 4
−1 4
)+…+(1 7
−1 3n−2
)1 3n+1
=
lim n→∞
(1−1 3
)1 3n+1
=
lim n→∞
•1 3
=3n 3n+1
．1 3
故答案为
．1 3
答案解析：首先利用列项相消法求出数列的和，然后取极限即可得到答案．
考试点：极限及其运算；数列的求和．
知识点：本题考查了列项相消法求数列的前n项和，考查了数列极限的求法，是基础的运算题．