已知数列{an}的通项公式an=pn^2+qn,(p,q属于R,且p,q为常数)bn=an+1-an求证对任意实数pq数列{bn}是等差数列

Question

已知数列{an}的通项公式an=pn^2+qn,(p,q属于R,且p,q为常数)bn=an+1-an求证对任意实数pq数列{bn}是等差数列

Answer 1

an=pn²+qn
bn=a(n+1)-an=p(n+1)²+q(n+1)-pn²-qn=2pn+p+q
b(n+1)=2p(n+1)+p+q
b(n+1)-bn=2p(n+1)+p+q-2pn-p-q=2p,为定值.
a1=p+q
a2=4p+2q
b1=a2-a1=4p+2q-p-q=3p+q
数列{bn}是以3p+q为首项,2p为公差的等差数列.