a1=1 an+1=2an+n+1 bn=an+p*n+q 求 p q为何值时 {bn} 为等比数列

an+1=2an+n+1
a(n+1)+k(n+1)+p=2(an+kn+p)
2k-k=1
2p-p-k=1
k=1
p=2
a(n+1)+(n+1)+2=2*(an+n+2)
{an+n+2}为等比数列,首项a1+3=4
an+n+2=4*2^(n-1)=2^(n+1)
an=2^(n+1)-n-2
bn=an+p*n+q=2^(n+1)-n-2+p*n+q=2^(n+1)+(p-1)n+q-2
b(n-1)=2^n+(p-1)*(n-1)+q-2
q=bn/b(n-1)=[2^(n+1)+(p-1)n+q-2]/[2^n+(p-1)*(n-1)+q-2]
要为常数,只能
(p-1)n+q-2=0
即
p=1
q=2