已知数列an满足a（n+1）=2an+3n^2+4n+5,a1=1,求数列an的通项公式

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已知数列an满足a（n+1）=2an+3n^2+4n+5,a1=1,求数列an的通项公式
不要复制,要如何配凑的过程

a(n+1)=2an+3n^2+4n+5
let
a(n+1) + k1(n+1)^2+k2(n+1)+k3=2(an+k1n^2+k2n+k3)
coef.of n^2
k1=3
coef.of n
k2-2k1=4
k2=10
coef.of constant
k3-k1-k2=5
k3=18
ie
a(n+1)=2an+3n^2+4n+5
a(n+1) + 3(n+1)^2+10(n+1)+18=2(an+3n^2+10n+18)
=>{an+3n^2+10n+18}是等比数列,q=2
an+3n^2+10n+18 = 2^(n-1) .(a1+3+10+18)
=2^(n+4)
an = -(3n^2+10n+18) +2^(n+4)

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