数列{an}中,a1=1,an+1=-an+n^2,求{an}的通项及a2000.求详解,
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数列{an}中,a1=1,an+1=-an+n^2,求{an}的通项及a2000.求详解,
![数列{an}中,a1=1,an+1=-an+n^2,求{an}的通项及a2000.求详解,](/uploads/image/z/15162946-34-6.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1%3D1%2Can%2B1%3D-an%2Bn%5E2%2C%E6%B1%82%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%8F%8Aa2000.%E6%B1%82%E8%AF%A6%E8%A7%A3%2C)
a(n+1)=-an+n^2
let
a(n+1) +k1(n+1)^2+k2(n+1) + k3 = -( an +k1n^2+k2n+k3)
coef.of n^2
-2k1=1
k1 = -1/2
coef.of n
-2k2-2k1=0
-2k2+1=0
k2 =1/2
coef.of constant
-2k3 -k1-k2 =0
-2k3+1/2-1/2=0
k3=0
ie
a(n+1) -(1/2)(n+1)^2+(1/2)(n+1) = -( an -(1/2)n^2+(1/2)n)
[a(n+1) -(1/2)(n+1)^2+(1/2)(n+1)]/( an -(1/2)n^2+(1/2)n) =-1
( an -(1/2)n^2+(1/2)n)/( a1 -(1/2)+(1/2)) = (-1)^(n-1)
an -(1/2)n^2+(1/2)n = (-1)^(n-1)
an = (1/2)n^2 -(1/2)n + (-1)^(n-1)
a2000 = 1998999
再问: 谢谢,不过你写的那些符号是什么意思唉,我看不懂额,我是个高中生,麻烦能不能解释一下?~
再答: a(n+1)=-an+n^2 设 a(n+1) +k1(n+1)^2+k2(n+1) + k3 = -( an +k1n^2+k2n+k3) n^2 的系数 -2k1=1 k1 = -1/2 n的系数 -2k2-2k1=0 -2k2+1=0 k2 =1/2 常数的系数 -2k3 -k1-k2 =0 -2k3+1/2-1/2=0 k3=0 故得 a(n+1) -(1/2)(n+1)^2+(1/2)(n+1) = -( an -(1/2)n^2+(1/2)n) [a(n+1) -(1/2)(n+1)^2+(1/2)(n+1)]/( an -(1/2)n^2+(1/2)n) =-1 设 bn= an -(1/2)n^2+(1/2)n bn 是等比数列, q=-1 b(n+1)/bn =-1 bn/b1= (-1)^(n-1) bn =(-1)^(n-1) an -(1/2)n^2+(1/2)n = (-1)^(n-1) an = (1/2)n^2 -(1/2)n + (-1)^(n-1) a2000 = 1998999
再问: 设a(n+1) +k1(n+1)^2+k2(n+1) + k3 = -( an +k1n^2+k2n+k3) 这一步是什么意思,哪儿得到的?是固定用法吗?我就这里看不懂额,麻烦你解释一下,谢谢!~
再答: 把原有的 a(n+1)=-an+n^2 变成这样 a(n+1) +k1(n+1)^2+k2(n+1) + k3 = -(an +k1n^2+k2n+k3) 那 bn=an +k1n^2+k2n+k3 bn就是一个等比数列 然后根据 a(n+1)=-an+n^2 再计算出k1, k2,k3 的值
let
a(n+1) +k1(n+1)^2+k2(n+1) + k3 = -( an +k1n^2+k2n+k3)
coef.of n^2
-2k1=1
k1 = -1/2
coef.of n
-2k2-2k1=0
-2k2+1=0
k2 =1/2
coef.of constant
-2k3 -k1-k2 =0
-2k3+1/2-1/2=0
k3=0
ie
a(n+1) -(1/2)(n+1)^2+(1/2)(n+1) = -( an -(1/2)n^2+(1/2)n)
[a(n+1) -(1/2)(n+1)^2+(1/2)(n+1)]/( an -(1/2)n^2+(1/2)n) =-1
( an -(1/2)n^2+(1/2)n)/( a1 -(1/2)+(1/2)) = (-1)^(n-1)
an -(1/2)n^2+(1/2)n = (-1)^(n-1)
an = (1/2)n^2 -(1/2)n + (-1)^(n-1)
a2000 = 1998999
再问: 谢谢,不过你写的那些符号是什么意思唉,我看不懂额,我是个高中生,麻烦能不能解释一下?~
再答: a(n+1)=-an+n^2 设 a(n+1) +k1(n+1)^2+k2(n+1) + k3 = -( an +k1n^2+k2n+k3) n^2 的系数 -2k1=1 k1 = -1/2 n的系数 -2k2-2k1=0 -2k2+1=0 k2 =1/2 常数的系数 -2k3 -k1-k2 =0 -2k3+1/2-1/2=0 k3=0 故得 a(n+1) -(1/2)(n+1)^2+(1/2)(n+1) = -( an -(1/2)n^2+(1/2)n) [a(n+1) -(1/2)(n+1)^2+(1/2)(n+1)]/( an -(1/2)n^2+(1/2)n) =-1 设 bn= an -(1/2)n^2+(1/2)n bn 是等比数列, q=-1 b(n+1)/bn =-1 bn/b1= (-1)^(n-1) bn =(-1)^(n-1) an -(1/2)n^2+(1/2)n = (-1)^(n-1) an = (1/2)n^2 -(1/2)n + (-1)^(n-1) a2000 = 1998999
再问: 设a(n+1) +k1(n+1)^2+k2(n+1) + k3 = -( an +k1n^2+k2n+k3) 这一步是什么意思,哪儿得到的?是固定用法吗?我就这里看不懂额,麻烦你解释一下,谢谢!~
再答: 把原有的 a(n+1)=-an+n^2 变成这样 a(n+1) +k1(n+1)^2+k2(n+1) + k3 = -(an +k1n^2+k2n+k3) 那 bn=an +k1n^2+k2n+k3 bn就是一个等比数列 然后根据 a(n+1)=-an+n^2 再计算出k1, k2,k3 的值
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