【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
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 05:11:18
![【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](/uploads/image/z/7274180-20-0.jpg?t=%E3%80%90%E9%AB%98%E8%80%83%E3%80%91%E6%95%B0%E5%88%97%E9%9A%BE%E9%A2%98%E5%8F%AF%E4%BB%A5%E8%AF%81%E6%98%8E%2C%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84n%E5%B1%9E%E4%BA%8EN%2B%2C%E6%9C%89%281%2B2%2B%E2%80%A6%E2%80%A6%2Bn%29%5E2%3D1%5E3%2B2%5E3%2B%E2%80%A6%E2%80%A6n%5E3%E6%88%90%E7%AB%8B%2C%E4%B8%8B%E9%9D%A2%E5%B0%9D%E8%AF%95%E6%8E%A8%E5%B9%BF%E8%AF%A5%E5%91%BD%E9%A2%98%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E6%AF%8F%E9%A1%B9%E5%9D%87%E9%9D%9E%E9%9B%B6%2C%E4%B8%94%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84n%E5%B1%9E%E4%BA%8EN%2B%E6%9C%89%28a1%2Ba2%2B%E2%80%A6%E2%80%A6%2Ban%29%5E2%3Da1%5E3%2Ba2%5E3%2B%E2%80%A6%E2%80%A6an%5E3%E6%88%90%E7%AB%8B%2C%E8%AF%95%E6%89%BE%E5%87%BA%E4%B8%80%E4%B8%AA)
【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
【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?
【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
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n2011时,a(n)=-a(n-1),
则有,a(2012)=-a(2011)=-2011.