一道数列求和已知a1 = a ,a2 = b (a,b 为常数)并且a3 = a1 + a2 ,a4 = a2 + a3 ,...an = a(n-1)+ a(n-2)求an 的表达式
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![一道数列求和已知a1 = a ,a2 = b (a,b 为常数)并且a3 = a1 + a2 ,a4 = a2 + a3 ,...an = a(n-1)+ a(n-2)求an 的表达式](/uploads/image/z/10190158-70-8.jpg?t=%E4%B8%80%E9%81%93%E6%95%B0%E5%88%97%E6%B1%82%E5%92%8C%E5%B7%B2%E7%9F%A5a1+%3D+a+%2Ca2+%3D+b+%28a%2Cb+%E4%B8%BA%E5%B8%B8%E6%95%B0%29%E5%B9%B6%E4%B8%94a3+%3D+a1+%2B+a2+%2Ca4+%3D+a2+%2B+a3+%2C...an+%3D+a%28n-1%EF%BC%89%2B+a%28n-2%29%E6%B1%82an+%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F)
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一道数列求和已知a1 = a ,a2 = b (a,b 为常数)并且a3 = a1 + a2 ,a4 = a2 + a3 ,...an = a(n-1)+ a(n-2)求an 的表达式
一道数列求和
已知
a1 = a ,a2 = b (a,b 为常数)
并且
a3 = a1 + a2 ,a4 = a2 + a3 ,...an = a(n-1)+ a(n-2)
求an 的表达式
一道数列求和已知a1 = a ,a2 = b (a,b 为常数)并且a3 = a1 + a2 ,a4 = a2 + a3 ,...an = a(n-1)+ a(n-2)求an 的表达式
你可以在百度里查一下Fibonacci数列.
an=1/√[(1+√5/2) n-(1-√5/2) n] (n=1,2,3.....)