题目内容

计算:(1+
1
2
+
1
3
+
1
4
+…+
1
201404
2+(
1
2
+
1
3
+
1
4
+…+
1
201404
2+(
1
3
+
1
4
+…+
1
201404
2+…+(
1
201404
2+(1+
1
2
+
1
3
+
1
4
+…+
1
201404
).
考点:数列的求和
专题:等差数列与等比数列
分析:由12+1=2×1;(1+
1
2
)2
+(
1
2
)2
+(1+
1
2
)
=4=2×2;(1+
1
2
+
1
3
)2
+(
1
2
+
1
3
)2
+(
1
3
)2
+(1+
1
2
+
1
3
)
=6=2×3;…,观察分析猜想归纳即可得出.
解答: 解:∵12+1=2×1;
(1+
1
2
)2
+(
1
2
)2
+(1+
1
2
)
=4=2×2;
(1+
1
2
+
1
3
)2
+(
1
2
+
1
3
)2
+(
1
3
)2
+(1+
1
2
+
1
3
)
=6=2×3;
…,
可得(1+
1
2
+
1
3
+
1
4
+…+
1
201404
2+(
1
2
+
1
3
+
1
4
+…+
1
201404
2+(
1
3
+
1
4
+…+
1
201404
2+…+(
1
201404
2+(1+
1
2
+
1
3
+
1
4
+…+
1
201404
)=2×201404=402808.
点评:本题考查了观察分析猜想归纳求数列的和,考查了推理能力与计算能力,属于中档题.
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