已知數(shù)列{an}的奇數(shù)項(xiàng)是公差為d1的等差數(shù)列,偶數(shù)項(xiàng)是公差為d2的等差數(shù)列,Sn是數(shù)列{an}的前n項(xiàng)和,a1=1,a2=2.
(1)若S5=16,a4=a5,求a10;
(2)已知S15=15a8,且對(duì)任意n∈N*,有an<an+1恒成立,求證:數(shù)列{an}是等差數(shù)列;
(3)若d1=3d2(d1≠0),且存在正整數(shù)m、n(m≠n),使得am=an.求當(dāng)d1最大時(shí),數(shù)列{an}的通項(xiàng)公式.
(1)解:根據(jù)題意,有a
1=1,a
2=2,a
3=a
1+d
1=1+d
1,a
4=a
2+d
2=2+d
2,a
5=a
3+d
1=1+2d
1
∵S
5=16,a
4=a
5,
∴a
1+a
2+a
3+a
4+a
5=7+3d
1+d
2=16,2+d
2=1+2d
1
∴d
1=2,d
2=3.
∴a
10=2+4d
2=14
(2)證明:當(dāng)n為偶數(shù)時(shí),∵a
n<a
n+1恒成立,∴2+
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,
∴
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(d
2-d
1)+1-d
2<0
∴d
2-d
1≤0且d
2>1
當(dāng)n為奇數(shù)時(shí),∵a
n<a
n+1恒成立,∴
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,
∴(1-n)(d
1-d
2)+2>0
∴d
1-d
2≤0
∴d
1=d
2∵S
15=15a
8,∴8+
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+14+
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=30+45d
2∴d
1=d
2=2
∴a
n=n
∴數(shù)列{a
n}是等差數(shù)列;
(3)解:若d
1=3d
2(d
1≠0),且存在正整數(shù)m、n(m≠n),使得a
m=a
n,在m,n中必然一個(gè)是奇數(shù),一個(gè)是偶數(shù)
不妨設(shè)m為奇數(shù),n為偶數(shù)
∵a
m=a
n,∴
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∵d
1=3d
2,∴
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∵m為奇數(shù),n為偶數(shù),∴3m-n-1的最小正值為2,此時(shí)d
1=3,d
2=1
∴數(shù)列{a
n}的通項(xiàng)公式為a
n=
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.
分析:(1)確定數(shù)列的前5項(xiàng),利用S
5=16,a
4=a
5,建立方程,求出d
1=2,d
2=3,從而可求a
10;
(2)先證明d
1=d
2,再利用S
15=15a
8,求得d
1=d
2=2,從而可證數(shù)列{a
n}是等差數(shù)列;
(3)若d
1=3d
2(d
1≠0),且存在正整數(shù)m、n(m≠n),使得a
m=a
n,在m,n中必然一個(gè)是奇數(shù),一個(gè)是偶數(shù).不妨設(shè)m為奇數(shù),n為偶數(shù),利用a
m=a
n,及d
1=3d
2,可得
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,從而可求當(dāng)d
1最大時(shí),數(shù)列{a
n}的通項(xiàng)公式.
點(diǎn)評(píng):本題考查數(shù)列的通項(xiàng),考查數(shù)列的求和,考查學(xué)生分析解決問(wèn)題的能力,確定數(shù)列的通項(xiàng)是關(guān)鍵.