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已知常数p>0,数列{an}满足an+1=|p-an|+2an+p,n∈N*.(1)若a1=-1,p=1,①求a4的值;②求数列{an}的前n项和Sn;(2)若数列{an}中存在三项ar,as,at(r,s,t∈N*,r<s<t)依次成等差数列,求a
题目详情
已知常数p>0,数列{an}满足an+1=|p-an|+2an+p,n∈N*.
(1)若a1=-1,p=1,
①求a4的值;
②求数列{an}的前n项和Sn;
(2)若数列{an}中存在三项ar,as,at(r,s,t∈N*,r<s<t)依次成等差数列,求
的取值范围.
(1)若a1=-1,p=1,
①求a4的值;
②求数列{an}的前n项和Sn;
(2)若数列{an}中存在三项ar,as,at(r,s,t∈N*,r<s<t)依次成等差数列,求
a1 |
p |
▼优质解答
答案和解析
(1)①∵an+1=|p-an|+2an+p,
∴a2=|1-a1|+2a1+1=2-2+1=1,
a3=|1-a2|+2a2+1=0+2+1=3,
a4=|1-a3|+2a3+1=2+6+1=9,
②∵a2=1,an+1=|1-an|+2an+1,
∴当n≥2时,an≥1,
当n≥2时,an+1=-1+an+2an+1=3an,即从第二项起,数列{an}是以1为首项,以3为公比的等比数列,
∴数列{an}的前n项和Sn=a1+a2+a3+a4+…+an=-1+
=
×3n-1-
,(n≥2),
显然当n=1时,上式也成立,
∴Sn=
×3n-1-
;
(2)∵an+1-an=|p-an|+an+p≥p-an+an+p=2p>0,
∴an+1>an,即{an}单调递增.
(i)当
≥1时,有a1≥p,于是an≥a1≥p,
∴an+1=|p-an|+2an+p=an-p+2an+p=3an,∴an=3n-1•a1.
若数列{an}中存在三项ar,as,at(r,s,t∈N*,r<s<t)依次成等差数列,则有2as=ar+at,
即2×3s-1=3r-1+3t-1.(*)
∵s≤t-1,∴2×3s-1=
×3s<3t-1<3r-1+3t-1.因此(*)不成立.因此此时数列{an}中不存在三项ar,as,at(r,s,t∈N*,r<s<t)依次成等差数列.
(ii)当-1<
<1时,有-p<a1<p.此时a2=|P-a1|+2a1+p=p-a1+2a1+p=a1+2p>p.
于是当n≥2时,an≥a2>p.从而an+1=|p-an|+2an+p=an-p+2an+p=3an.∴an=3n-2a2=3n-2(a1+2p)(n≥2).
若数列{an}中存在三项ar,as,at(r,s,t∈N*,r<s<t)依次成等差数列,则有2as=ar+at,
同(i)可知:r=1.于是有2×3s-2(a1+2p)=a1+3t-2(a1+2p),∵2≤S≤t-1,∴
=2×3s-2-3t-2=
×3s-
×3t-1<0.∵2×3s-2-3t-2是整数,∴
≤-1.于是a1≤-a1-2p,即a1≤-p.与-p<a1<p矛盾.
故此时数列{an}中不存在三项ar,as,at(r,s,t∈N*,r<s<t)依次成等差数列.
(iii)当
≤-1时,有a1≤-p<p.a1+p≤0.
于是a2=|P-a1|+2a1+p=p-a1+2a1+p=a1+2p.
a3=|p-a2|+2a2+p=|a1+p|+2a1+5p.=-a1-p+2a1+5p=a1+4p.
此时数列{an}中存在三项a1,a2,a3依次成等差数列.
综上可得:
∴a2=|1-a1|+2a1+1=2-2+1=1,
a3=|1-a2|+2a2+1=0+2+1=3,
a4=|1-a3|+2a3+1=2+6+1=9,
②∵a2=1,an+1=|1-an|+2an+1,
∴当n≥2时,an≥1,
当n≥2时,an+1=-1+an+2an+1=3an,即从第二项起,数列{an}是以1为首项,以3为公比的等比数列,
∴数列{an}的前n项和Sn=a1+a2+a3+a4+…+an=-1+
1-3n-1 |
1-3 |
1 |
2 |
3 |
2 |
显然当n=1时,上式也成立,
∴Sn=
1 |
2 |
3 |
2 |
(2)∵an+1-an=|p-an|+an+p≥p-an+an+p=2p>0,
∴an+1>an,即{an}单调递增.
(i)当
a1 |
p |
∴an+1=|p-an|+2an+p=an-p+2an+p=3an,∴an=3n-1•a1.
若数列{an}中存在三项ar,as,at(r,s,t∈N*,r<s<t)依次成等差数列,则有2as=ar+at,
即2×3s-1=3r-1+3t-1.(*)
∵s≤t-1,∴2×3s-1=
2 |
3 |
(ii)当-1<
a1 |
p |
于是当n≥2时,an≥a2>p.从而an+1=|p-an|+2an+p=an-p+2an+p=3an.∴an=3n-2a2=3n-2(a1+2p)(n≥2).
若数列{an}中存在三项ar,as,at(r,s,t∈N*,r<s<t)依次成等差数列,则有2as=ar+at,
同(i)可知:r=1.于是有2×3s-2(a1+2p)=a1+3t-2(a1+2p),∵2≤S≤t-1,∴
a1 |
a1+2p |
2 |
9 |
1 |
3 |
a1 |
a1+2p |
故此时数列{an}中不存在三项ar,as,at(r,s,t∈N*,r<s<t)依次成等差数列.
(iii)当
a1 |
p |
于是a2=|P-a1|+2a1+p=p-a1+2a1+p=a1+2p.
a3=|p-a2|+2a2+p=|a1+p|+2a1+5p.=-a1-p+2a1+5p=a1+4p.
此时数列{an}中存在三项a1,a2,a3依次成等差数列.
综上可得:
a1 |
p |
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