是否存在常數(shù)a,b使等式1-n+2-(n-1)+3-(n-2)+…+n-1=an(n+b)(n+2)對于任意的n∈N+總成立?若存在,求出來并證明;若不存在,說明理由.
【答案】
分析:可假設(shè)存在常數(shù)a,b使等式1-n+2-(n-1)+3-(n-2)+…+n-1=an(n+b)(n+2)對于任意的n∈N
+總成立,令n=1與n=2列方程解得a,b再用數(shù)學(xué)歸納法證明.
解答:解:假設(shè)存在常數(shù)a,b使等式1-n+2-(n-1)+3-(n-2)+…+n-1=an(n+b)(n+2)對于任意的n∈N
+總成立,
令n=1與n=2得:
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解得:
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,
即1-n+2-(n-1)+3-(n-2)+…+n-1=
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n(n+1)(n+2).
下面用數(shù)學(xué)歸納法證明:
(1)當(dāng)n=1時,左邊=1×1=1,右邊=
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×1×1×(1+1)×(1+2)=1,因此左邊=右邊,
∴當(dāng)n=1時等式成立,
(2)假設(shè)當(dāng)n=k時成立,
即1×k+2×(k-1)+3×(k-2)+…+k×1=
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k(k+1)(k+2),
那么當(dāng) n=k+1時,
1×(k+1)+2×[(k+1)-1]+3×[(k+1)-2)]+…+(k+1)×1
=[1×k+2×(k-1)+3×(k-2)+…+k×1]+[1+2+3+…+(k+1)]
=
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k(k+1)(k+2)+
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=
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(k+1)(k+2)(k+3)
=
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(k+1)[(k+1)+1][(k+1)+2]
所以,當(dāng) n=k+1時等式也成立.
根據(jù)(1)和(2),可知等式對任何n∈N
+都成立.
點評:本題考查數(shù)學(xué)歸納法,對于本題“是否存在”型的問題,先假設(shè)存在,通過題意求得a、b的值,再用數(shù)學(xué)歸納法予以證明,難點在于n=k+1時,等式成立的證明,要用好歸納假設(shè),屬于中檔題.