【答案】
分析:(1)根據(jù)a
n=p
n+λq
n可得a
n+1-pa
n的表達(dá)式,整理可得
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為常數(shù),進(jìn)而可判斷數(shù)列{a
n+1-pa
n}為等比數(shù)列.
(2)取數(shù)列{a
n}的連續(xù)三項(xiàng)a
n,a
n+1,a
n+2把a(bǔ)
n=p
n+λq
n代入a
n+12-a
na
n+2整理可知結(jié)果不為0,進(jìn)而可判斷a
n+12≠a
na
n+2,即數(shù)列{a
n}中不存在連續(xù)三項(xiàng)構(gòu)成等比數(shù)列;
(3)由3
n+2
n=5
n整理得
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,設(shè)
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則可知f(x)為減函數(shù),故可判定f(x)=1的解只有一個(gè),從而當(dāng)且僅當(dāng)n=1,3
n+2
n=5
n成立,同樣的道理可證當(dāng)k=1,k=3或k≥5時(shí),B∩C=∅;當(dāng)k=2時(shí),B∩C={(1,5)},當(dāng)k=4時(shí),B∩C={(2,25)}.
解答:解:(1)∵a
n=p
n+λq
n,
∴a
n+1-pa
n=p
n+1+λq
n+1-p(p
n+λq
n)=λq
n(q-p),
∵λ≠0,q>0,p≠q
∴
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為常數(shù)
∴數(shù)列{a
n+1-pa
n}為等比數(shù)列
(2)取數(shù)列{a
n}的連續(xù)三項(xiàng)a
n,a
n+1,a
n+2(n≥1,n∈N
*),
∵a
n+12-a
na
n+2=(p
n+1+λq
n+1)
2-(p
n+λq
n)(p
n+2+λq
n+2)=-λp
nq
n(p-q)
2,
∵p>0,q>0,p≠q,λ≠0,
∴-λp
nq
n(p-q)
2≠0,即a
n+12≠a
na
n+2,
∴數(shù)列{a
n}中不存在連續(xù)三項(xiàng)構(gòu)成等比數(shù)列;
(3)當(dāng)k=1時(shí),3
n+k
n=3
n+1<5
n,此時(shí)B∩C=∅;
當(dāng)k=3時(shí),3
n+k
n=3
n+3
n=2•3
n為偶數(shù);而5
n為奇數(shù),此時(shí)B∩C=∅;
當(dāng)k≥5時(shí),3
n+k
n>5
n,此時(shí)B∩C=∅;
當(dāng)k=2時(shí),3
n+2
n=5
n,發(fā)現(xiàn)n=1符合要求,
下面證明唯一性(即只有n=1符合要求).
由3
n+2
n=5
n得
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,
設(shè)
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,則
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是R上的減函數(shù),
∴f(x)=1的解只有一個(gè)
從而當(dāng)且僅當(dāng)n=1時(shí)

,
即3
n+2
n=5
n,此時(shí)B∩C={(1,5)};
當(dāng)k=4時(shí),3
n+4
n=5
n,發(fā)現(xiàn)n=2符合要求,
下面同理可證明唯一性(即只有n=2符合要求).
從而當(dāng)且僅當(dāng)n=2時(shí)
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,
即3
n+4
n=5
n,此時(shí)B∩C={(2,25)};
綜上,當(dāng)k=1,k=3或k≥5時(shí),B∩C=∅;
當(dāng)k=2時(shí),B∩C={(1,5)},
當(dāng)k=4時(shí),B∩C={(2,25)}.
點(diǎn)評(píng):本題主要考查了等比數(shù)列的確定和集合的相關(guān)知識(shí).考查了學(xué)生分析和運(yùn)算能力.