解:(1)第1组3=2×1+1,
第2组5=2×2+1,
第3组7=2×3+1,
…
第n组有2n+1粒,
所以第100组应该有种子2×100+1=201粒.
(2)由a
1=
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+
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=
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=
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,
a
2=
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+
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=
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=
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,
a
3=
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+
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=
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=
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,
…
所以a
99=
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+
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=
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=
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;
(3)第1个图案由4=4×1-(1-1)=3+1个基础图形组成,
第2个图案由7=4×2-(2-1)=3×2+1个基础图形组成,
第3个图案由10=4×3-(3-1)=3×3+1个基础图形组成,
…,
那么第101个图案中由3×101+1=304个基础图形组成;
(4)把
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,
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,
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…代入得:
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=1-
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+
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-
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+
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-
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+…+
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-
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=1-
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=
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.
分析:(1)第1组取3粒,第2组取5粒,第3组取7粒…第n组取(2n-1)粒解决问题;
(2)所有第一个加数分子是1,分母是三个连续自然数的乘积,第二个加数分子是1,分母是三个连续自然数中间的数,和的分母是三个连续自℃然数两端数的乘积,分子是三个连续自然数中间的数,因此可求得a
99.
(3)第1个图案由4个基础图形组成,第2个图案由7个基础图形组成,第3个图案由10个基础图形组成,…,第n个图案由(3n+1)个基础图形组成,由此解决问题;
(4)由
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,
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,
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,…
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=
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,代入可解决问题.
点评:抓住式子的变化规律或数的变化规律,就可以解决问题.