已知{an}为等差数列,且an≠0,公差d≠0(1)数列满足结论1/a1-1/a2=d/a1a2;1/a1-2/a2+1/a3=2d²/a1a2a3;试证:1/a1-3/a2+3/a3-1/a4=6d³/a1a2a3a4(2)根据(1)中的几个等式,试归纳出更一般的结论,并用数学
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![已知{an}为等差数列,且an≠0,公差d≠0(1)数列满足结论1/a1-1/a2=d/a1a2;1/a1-2/a2+1/a3=2d²/a1a2a3;试证:1/a1-3/a2+3/a3-1/a4=6d³/a1a2a3a4(2)根据(1)中的几个等式,试归纳出更一般的结论,并用数学](/uploads/image/z/5479276-4-6.jpg?t=%E5%B7%B2%E7%9F%A5%7Ban%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E4%B8%94an%E2%89%A00%2C%E5%85%AC%E5%B7%AEd%E2%89%A00%EF%BC%881%EF%BC%89%E6%95%B0%E5%88%97%E6%BB%A1%E8%B6%B3%E7%BB%93%E8%AE%BA1%2Fa1-1%2Fa2%3Dd%2Fa1a2%3B1%2Fa1-2%2Fa2%2B1%2Fa3%3D2d%26%23178%3B%2Fa1a2a3%3B%E8%AF%95%E8%AF%81%EF%BC%9A1%2Fa1-3%2Fa2%2B3%2Fa3-1%2Fa4%3D6d%26%23179%3B%2Fa1a2a3a4%EF%BC%882%EF%BC%89%E6%A0%B9%E6%8D%AE%EF%BC%881%EF%BC%89%E4%B8%AD%E7%9A%84%E5%87%A0%E4%B8%AA%E7%AD%89%E5%BC%8F%2C%E8%AF%95%E5%BD%92%E7%BA%B3%E5%87%BA%E6%9B%B4%E4%B8%80%E8%88%AC%E7%9A%84%E7%BB%93%E8%AE%BA%2C%E5%B9%B6%E7%94%A8%E6%95%B0%E5%AD%A6)
已知{an}为等差数列,且an≠0,公差d≠0(1)数列满足结论1/a1-1/a2=d/a1a2;1/a1-2/a2+1/a3=2d²/a1a2a3;试证:1/a1-3/a2+3/a3-1/a4=6d³/a1a2a3a4(2)根据(1)中的几个等式,试归纳出更一般的结论,并用数学
已知{an}为等差数列,且an≠0,公差d≠0
(1)数列满足结论1/a1-1/a2=d/a1a2;1/a1-2/a2+1/a3=2d²/a1a2a3;试证:1/a1-3/a2+3/a3-1/a4=6d³/a1a2a3a4
(2)根据(1)中的几个等式,试归纳出更一般的结论,并用数学归纳法证明
已知{an}为等差数列,且an≠0,公差d≠0(1)数列满足结论1/a1-1/a2=d/a1a2;1/a1-2/a2+1/a3=2d²/a1a2a3;试证:1/a1-3/a2+3/a3-1/a4=6d³/a1a2a3a4(2)根据(1)中的几个等式,试归纳出更一般的结论,并用数学
哎我也是新海的 高二周练伤不起
解析没怎么看懂 坑
提问者是我大新海的吗!
证明 (1)∵{an}是等差数列,∴2a(k+1)=a(k)+a(k+2),
故方程a(k)x^2+2a(k+1)x+a(k+2)=0可变为[a(k)x+a(k+2)](x+1)=0,
∴当k取不同自然数时,原方程有一个公共根-1
(2)原方程不同的根为x(k)=-[a(k+2)]/a(k)=-[a(k)+2d]/a(k)=-1-[2d/a(k)]
∴1/...
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证明 (1)∵{an}是等差数列,∴2a(k+1)=a(k)+a(k+2),
故方程a(k)x^2+2a(k+1)x+a(k+2)=0可变为[a(k)x+a(k+2)](x+1)=0,
∴当k取不同自然数时,原方程有一个公共根-1
(2)原方程不同的根为x(k)=-[a(k+2)]/a(k)=-[a(k)+2d]/a(k)=-1-[2d/a(k)]
∴1/[x(k)+1]=-[a(k)/2d]
∵1/[x(k+1)+1]-1/[x(k)+1]=-[a(k+1)/2d]-[-a(k)/2d]=[a(k)-a(k+1)]/2d=-d/2d=-1/2
∴{1/x(k+1)}是以-1/2为公差的等差数列
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