已知{An}成等差数列,Bn=[A(n+1)]^2-An^2,求证数列{Bn}也成等差数列.
答案:2 悬赏:60 手机版
解决时间 2021-03-23 06:01
- 提问者网友：低吟詩仙的傷
- 2021-03-22 17:39
已知{An}成等差数列,Bn=[A(n+1)]^2-An^2,求证数列{Bn}也成等差数列.
最佳答案
- 五星知识达人网友：第四晚心情
- 2021-03-22 18:57
B(n+1)+B(n-1)=A(n+2)^2-A(n+1)^2+An^2-A(n-1)^2
=A(n+2)^2-A(n-1)^2
=(A(n+2)-A(n-1))(A(n+2)+A(n-1))
因为An是等差数列
=2d[A(n+2)+A(n-1)]
2Bn=2A(n+1)^2-2An^2
=2[A(n+1)-An][A(n+1)+An]
=2d[A(n+1)+An]
因为A(n+2)+A(n-1)=A(n+1)+An
所以B(n+1)+B(n-1)=2Bn
所以Bn是等差数列
=A(n+2)^2-A(n-1)^2
=(A(n+2)-A(n-1))(A(n+2)+A(n-1))
因为An是等差数列
=2d[A(n+2)+A(n-1)]
2Bn=2A(n+1)^2-2An^2
=2[A(n+1)-An][A(n+1)+An]
=2d[A(n+1)+An]
因为A(n+2)+A(n-1)=A(n+1)+An
所以B(n+1)+B(n-1)=2Bn
所以Bn是等差数列
全部回答
- 1楼网友：琴狂剑也妄
- 2021-03-22 19:23
设等差数列｛an｝的公差为d,
bn-b(n-1)=a(n+1)^2-an^-[an^2-a(n-1)^2]
=[a(n+1)+an]*[a(n+1)-an] - [an+a(n-1)]*[an-a(n-1)]
=d[a(n+1)+an] - d[an+a(n-1)]
=d[a(n+1)+an-an-a(n-1)]
=d[a(n+1)-a(n-1)]
=d*2d
=2d^2
我要举报
如以上回答内容为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！
点此我要举报以上问答信息
大家都在看
推荐资讯