题目内容
计算:(1+
+
+
+…+
)2+(
+
+
+…+
)2+(
+
+…+
)2+…+(
)2+(1+
+
+
+…+
).
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| 201404 |
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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+
)2+(
)2+(1+
)=4=2×2;(1+
+
)2+(
+
)2+(
)2+(1+
+
)=6=2×3;…,观察分析猜想归纳即可得出.
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解答:
解:∵12+1=2×1;
(1+
)2+(
)2+(1+
)=4=2×2;
(1+
+
)2+(
+
)2+(
)2+(1+
+
)=6=2×3;
…,
可得(1+
+
+
+…+
)2+(
+
+
+…+
)2+(
+
+…+
)2+…+(
)2+(1+
+
+
+…+
)=2×201404=402808.
(1+
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(1+
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…,
可得(1+
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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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