已知数列{an}满足A1=0,A(N+1)=AN减去根号三/根号三AN加1.求A2,A3,A4,A5,A6N为正整数
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![已知数列{an}满足A1=0,A(N+1)=AN减去根号三/根号三AN加1.求A2,A3,A4,A5,A6N为正整数](/uploads/image/z/3980339-35-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3A1%3D0%2CA%28N%2B1%29%3DAN%E5%87%8F%E5%8E%BB%E6%A0%B9%E5%8F%B7%E4%B8%89%2F%E6%A0%B9%E5%8F%B7%E4%B8%89AN%E5%8A%A01.%E6%B1%82A2%2CA3%2CA4%2CA5%2CA6N%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0)
已知数列{an}满足A1=0,A(N+1)=AN减去根号三/根号三AN加1.求A2,A3,A4,A5,A6N为正整数
已知数列{an}满足A1=0,A(N+1)=AN减去根号三/根号三AN加1.求A2,A3,A4,A5,A6
N为正整数
已知数列{an}满足A1=0,A(N+1)=AN减去根号三/根号三AN加1.求A2,A3,A4,A5,A6N为正整数
A(n+1)=An-√3/(√3+1)
=An-√3(√3-1)/2
=An-3/2+√3/2
A2=3/2-√3/2
A3=A2-(-3/2+√3/2)=3-√3
A4=A3-(-3/2+√3/2)=9/2-3√3/2
A5=6-2√3
A6=15/2-5√3/2
首先二次根号的表示为sqrt 再次你能不能加几个括号啊?
这样没法理解!!!!!!!!!
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