题目内容

1 | ||
(n+1)
|
1 |
3 |
1 |
12 |
1 |
30 |
1 |
60 |
1 | ||
n
|
1 | ||
(n+1)
|
观察莱布尼兹三角形规律,计算极限
lim |
n→∞ |
1 |
2 |
1 |
2 |
分析:由于an=
+
+
+
+…+
+
=
+
+
+
=(1×
-
×
)+(
×
-
×
)+…+
-
),利用分组裂项求和,然后对其求极限
an=
(
-
)即可
1 |
3 |
1 |
12 |
1 |
30 |
1 |
60 |
1 | ||
n
|
1 | ||
(n+1)
|
2 |
1×2×3 |
2 |
2×3×4 |
2 |
3×4×5 |
2 |
n(n-1)(n+1) |
=(1×
1 |
2 |
1 |
2 |
1 |
3 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
(n(n-1) |
1 |
n(n+1) |
lim |
n→∞ |
lim |
n→∞ |
1 |
2 |
1 |
n(n+1) |
解答:解:∵an=
+
+
+
+…+
+
=
+
+
+
=(1×
-
×
)+(
×
-
×
)+…+
-
)
=(1-
+
-
+…+
-
)-(
-
+
-
+…+
-
)
=1-
-(
-
)
=
+
-
=
-
∴
an=
(
-
)=
故答案为:
1 |
3 |
1 |
12 |
1 |
30 |
1 |
60 |
1 | ||
n
|
1 | ||
(n+1)
|
=
2 |
1×2×3 |
2 |
2×3×4 |
2 |
3×4×5 |
2 |
n(n-1)(n+1) |
=(1×
1 |
2 |
1 |
2 |
1 |
3 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
(n(n-1) |
1 |
n(n+1) |
=(1-
1 |
2 |
1 |
2 |
1 |
3 |
1 |
n-1 |
1 |
n |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
n |
1 |
n+1 |
=1-
1 |
n |
1 |
2 |
1 |
n+1 |
=
1 |
2 |
1 |
n+1 |
1 |
n |
1 |
2 |
1 |
n(n+1) |
∴
lim |
n→∞ |
lim |
n→∞ |
1 |
2 |
1 |
n(n+1) |
1 |
2 |
故答案为:
1 |
2 |
点评:本题主要考查了数列极限的求解,解题的关键是根据题中给出的数列的通项公式,发现其裂项的规律,进而利用分组列项进行求解.

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