已知等差数列的前n项和Sn=pn^2-2n q(p,q∈R,n∈R).求q的值

题目

已知等差数列的前n项和Sn=pn^2-2n q(p,q∈R,n∈R).求q的值

答案

Sn=pn^2-2n +q
an=Sn-S(n-1)=pn^2-2n+q-[p(n-1)^2-2(n-1)+q]
=p[n^2-(n-1)^2]-2n+2(n-1)
=p(2n-1)-2
=2pn-p-2
公差d=an-a(n-1)=2pn-p-2-[2p(n-1)-p-2]=2p
a1=2p-p-2=p-2
Sn=na1+1/2n(n-1)d=n(p-2)+1/2n(n-1)*2p=np-2n+pn^2-pn
=pn^2-2n
=pn^2-2n+q(条件已知）
所以：q=0