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
问题描述：
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
答
an+1=2an+n+1a(n+1)+k(n+1)+p=2(an+kn+p)2k-k=12p-p-k=1k=1p=2a(n+1)+(n+1)+2=2*(an+n+2){an+n+2}为等比数列,首项a1+3=4an+n+2=4*2^(n-1)=2^(n+1)an=2^(n+1)-n-2bn=an+p*n+q=2^(n+1)-n-2+p*n+q=2^(n+1)+(p-1)n+q-2b(...