a1=1,a(n+1)=2an+n^2+1,求通项

Question

a1=1,a(n+1)=2an+n^2+1,求通项

Answer 1

因为a1=1,a(n+1)=2an+n^2+1
所以设a(n+1)+i*(n+1)^2+j*(n+1)+k=2*[an+i*n^2+j*n+k]
展开化简得a(n+1)=2an+i*n^2+(j-2i)*n+k-i-j
对比系数得i=1,j-2i=0,k-i-j=1
所以i=1,j=2,k=4
故a(n+1)+(n+1)^2+2(n+1)+4=2*[an+n^2+2n+4]
所以数列{an+n^2+2n+4}是等比数列,公比是q=2,首项是a1+1^2+2*1+4=8
所以an+n^2+2n+4=8*2^(n-1)=2^(n+2)
那么an=2^(n+2)-n^2-2n-4
如果不懂,请Hi我,祝学习愉快!