A(n+1)=2A(n)+n^2+2n+1求数列An通项公式
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a(n+1)=2an+n^2+2n+1
a(n+1)+k1(n+1)^2 + k2(n+1) + k3 = 2(an + k1n^2+k2n+k3)
coef. of n^2
k1=1
coef. of n
k2-2k1=2
k2=4
coef. of constant
k3-k2-k1=1
k3-4-1=1
k3=6
ie
a(n+1)+(n+1)^2 + 4(n+1) + 6 = 2(an + n^2+4n+6)
[a(n+1)+(n+1)^2 + 4(n+1) + 6]/ (an + n^2+4n+6) =2
(an + n^2+4n+6)/(a1+11)=2^(n-1)
an = (a1+11) . 2^(n-1) - (n^2+4n+6)
a(n+1)+k1(n+1)^2 + k2(n+1) + k3 = 2(an + k1n^2+k2n+k3)
coef. of n^2
k1=1
coef. of n
k2-2k1=2
k2=4
coef. of constant
k3-k2-k1=1
k3-4-1=1
k3=6
ie
a(n+1)+(n+1)^2 + 4(n+1) + 6 = 2(an + n^2+4n+6)
[a(n+1)+(n+1)^2 + 4(n+1) + 6]/ (an + n^2+4n+6) =2
(an + n^2+4n+6)/(a1+11)=2^(n-1)
an = (a1+11) . 2^(n-1) - (n^2+4n+6)
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