已知数列{an}满足a1=t>1,a(n+1)=(n+1)an/n,函数f(x)=ln(1+x)+mx^2-x (m属于
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已知数列{an}满足a1=t>1,a(n+1)=(n+1)an/n,函数f(x)=ln(1+x)+mx^2-x (m属于[0,0.5]) (1)求an (2单调性
已知数列{an}满足a1=t>1,a(n+1)=(n+1)an/n,函数f(x)=ln(1+x)+mx^2-x (m属于[0,0.5]) (1)求an (2)讨论f(x)的单调性
已知数列{an}满足a1=t>1,a(n+1)=(n+1)an/n,函数f(x)=ln(1+x)+mx^2-x (m属于[0,0.5]) (1)求an (2)讨论f(x)的单调性
![已知数列{an}满足a1=t>1,a(n+1)=(n+1)an/n,函数f(x)=ln(1+x)+mx^2-x (m属于](/uploads/image/z/3874305-57-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%BB%A1%E8%B6%B3a1%3Dt%EF%BC%9E1%2Ca%28n%2B1%29%3D%28n%2B1%29an%2Fn%2C%E5%87%BD%E6%95%B0f%28x%29%3Dln%281%2Bx%29%2Bmx%5E2-x+%28m%E5%B1%9E%E4%BA%8E)
(1)a(n+1)/an=1+1/n,代入a1=t
则a2=2t a3=3t a4=4t
则{an}是一个以t为首项 公差为t的等差数列
an=nt(n+1)/2
(2)f'(x)=1/x+1+2mx-1
当f’(x)>0时为单调递增 所以f‘(x)=2mx2+2mx+1/x+1
解得:f(x)单增区间为[-1,+∞)
则a2=2t a3=3t a4=4t
则{an}是一个以t为首项 公差为t的等差数列
an=nt(n+1)/2
(2)f'(x)=1/x+1+2mx-1
当f’(x)>0时为单调递增 所以f‘(x)=2mx2+2mx+1/x+1
解得:f(x)单增区间为[-1,+∞)
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