a1=a2=1,an+2=3an+1+18an+2n次方求an

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/07/01 13:09:13

先求特解令 an*=b×2^n ,则 b×2^(n+2)=3×b×2^(n+1)+18×b×2^n+2^n
有 4b=6b+18b+1 得b=-1/20 即特解为an*=-2^n/20
令cn=an-an* 可化简为 c[n+2]=3c[n+1]+18c[n]
且知c[1]=a1-a1*=1+2/20=11/10 c[2]=a2-a2*=1+4/20=6/5
配等比数列有c[n+2]+3c[n+1]=6(c[n+1]+3c[n])=...=6^n (c[2]+3c[1])=(9/2)6^n
和c[n+2]-6c[n+1]=-3(c[n+1]-6c[n])=...=(-3)^n(c[2]-6c[1])=(-27/5)(-3)^n
两式消去c[n+2]可得到 9c[n+1]=(9/2)6^n+(27/5)(-3)^n
c[n+1]=(1/2)6^n+(3/5)(-3)^n
c[n]=(1/2)6^(n-1)+(3/5)(-3)^(n-1) n≥2时
n=1 (1/2)6^(n-1)+(3/5)(-3)^(n-1) =1/2+3/5=11/10=c[1]
所以 c[n]=(1/2)6^(n-1)+(3/5)(-3)^(n-1) n≥1时
所以 a[n]=(1/2)6^(n-1)+(3/5)(-3)^(n-1)-2^n/20=(1/12)6^n-(1/5)(-3)^n-2^n/20 n≥1时

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