【答案】
分析:(1)由題意可知
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,
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,將a
1=1代入上式可得a
2=2.
(2)由
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,得a
13+a
23++a
n3=(a
1+a
2++a
n)
2解得a
n+12-a
n2=a
n+1+a
n.
所以a
n+1-a
n=1.由此能夠?qū)С鯽
n=n.
(3)由于(1+x)
n=C
n+C
n1x+C
n2x
2+C
n3x
3+,(1-x)
n=C
n-C
n1x+C
n2x
2-C
n3x
3+,所以(1+x)
n-(1-x)
n-2nx=2C
n3x
3+2C
n5x
5+.再由分析法可知原不等式成立.
解答:(1)解:當(dāng)n=1時(shí),有
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,
由于a
n>0,所以a
1=1.
當(dāng)n=2時(shí),有
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,即
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,
將a
1=1代入上式,由于a
n>0,所以a
2=2.
(2)解:由
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,
得a
13+a
23++a
n3=(a
1+a
2++a
n)
2,①
則有a
13+a
23++a
n3+a
n+13=(a
1+a
2++a
n+a
n+1)
2.②
②-①,得a
n+13=(a
1+a
2++a
n+a
n+1)
2-(a
1+a
2++a
n)
2,
由于a
n>0,所以a
n+12=2(a
1+a
2++a
n)+a
n+1.③
同樣有a
n2=2(a
1+a
2++a
n-1)+a
n(n≥2),④
③-④,得a
n+12-a
n2=a
n+1+a
n.
所以a
n+1-a
n=1.
由于a
2-a
1=1,即當(dāng)n≥1時(shí)都有a
n+1-a
n=1,所以數(shù)列{a
n}是首項(xiàng)為1,公差為1的等差數(shù)列.
故a
n=n.
(3)證明1:由于(1+x)
n=C
n+C
n1x+C
n2x
2+C
n3x
3+,(1-x)
n=C
n-C
n1x+C
n2x
2-C
n3x
3+,
所以(1+x)
n-(1-x)
n=2C
n1x+2C
n3x
3+2C
n5x
5+.
即(1+x)
n-(1-x)
n-2nx=2C
n3x
3+2C
n5x
5+.
令
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,則有
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.
即
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,
即(2n+1)
n≥(2n)
n+(2n-1)
n故a
2n+1n≥a
2nn+a
2n-1n.
證明2:要證a
2n+1n≥a
2nn+a
2n-1n,
只需證(2n+1)
n≥(2n)
n+(2n-1)
n,
只需證
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,
只需證
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.
由于
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=
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=
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=
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.
因此原不等式成立.
點(diǎn)評(píng):本小題主要考查數(shù)列、不等式、二項(xiàng)式定理等知識(shí),考查化歸與轉(zhuǎn)化的數(shù)學(xué)思想方法,以及抽象概括能力、運(yùn)算求解能力和創(chuàng)新意識(shí).