Question

【高考】数列难题

可以证明,对任意的n属于N+,有(1+2+……+n)^2=1^3+2^3+……n^3成立,下面尝试推广该命题
设数列{an}每项均非零,且对任意的n属于N+有(a1+a2+……+an)^2=a1^3+a2^3+……an^3成立,试找出一个无穷数列{an},使得a2012=-2011,则这样的数列{an}的一个通项公式是?

Answer 1

a(n)非零,
[a(1)]^2=[a(1)]^3,1=a(1).
[a(n+1)]^3=[a(1)+a(2)+...+a(n+1)]^2-[a(1)+a(2)+...+a(n)]^2=a(n+1)[a(n+1)+2a(1)+2a(2)+...+2a(n)],
[a(n+1)]^2=a(n+1)+2a(1)+2a(2)+...+2a(n),
[a(2)]^2=a(2)+2a(1)=a(2)+2, 0=[a(2)]^2-a(n)-2=[a(2)-2][a(2)+1], a(2)=2或a(2)=-1.

[a(n+2)]^2=a(n+2)+2a(1)+2a(2)+...+2a(n)+2a(n+1),
[a(n+2)]^2-[a(n+1)]^2=a(n+2)-a(n+1)+2a(n+1)=a(n+2)+a(n+1),
0=[a(n+2)+a(n+1)][a(n+2)-a(n+1)-1],
a(n+2)=-a(n+1)或a(n+2)=a(n+1)+1,

让我们看看这种数列的各种可能的取值:
a(1)=1
a(2)=2,-1
a(3)=3,-2,1
a(4)=4,-3,2,-1,
a(5)=5,-4,3,-2,1,
...
a(2011)=2011,-2010,2009,...,-2,1,
a(2012)=2012,-2011,2010,...,2,-1
若n<=2011时,a(n)=n,
n>2011时,a(n)=-a(n-1),
则有,a(2012)=-a(2011)=-2011.

Answer 2

当x=1.时，An等于Bn
当x∈(1/10,1)，A3<B3,假设当n=k时，An<Bn
即(1+lgx)^k<1+klgx+k(k-1)/2(lgx)^2;
两边同时乘以(1+lgx)得
(1+lgx)^(k+1)<1+(k+1)lgx+k(k+1)/2(lgx)^2+k(k-1)/2*(lgx)^3
因为x∈(1/10,1),则lgx<0，故(1+lgx)^(k+1)<1+(k+1)lgx+k(k+1)/2(lgx)^2+k(k-1)/2*(lgx)^3<1+(k+1)lgx+k(k+1)/2(lgx)^2;
即当n=k+1,An<bn也成立,所以
当x∈(1/10,1)时，An<Bn。

当x∈(1,+∞).A3>B3.假设当n=k时，An>Bn,即(1+lgx)^k>1+klgx+k(k-1)/2(lgx)^2
两边同时乘以(1+lgx)得
(1+lgx)^(k+1)>1+(k+1)lgx+k(k+1)/2(lgx)^2+k(k-1)/2*(lgx)^3
因为lgx>0.所以(1+lgx)^(k+1)>1+(k+1)lgx+k(k+1)/2(lgx)^2+k(k-1)/2*(lgx)^3>1+(k+1)lgx+k(k+1)/2(lgx)^2,
即A[k+1]>B[k+1],n=k+1时也成立;
故当x∈(1,+∞)时，An>Bn。