Question

已知等差数列｛an｝的前n项和sn=tn^2+(t-9)n+t-3/2(t为常数)，求｛an｝的通项公式

Answer 1

n=1时,a1=S1=t+(t-9)+t-3/2=3t-21/2
n>1时,
an=Sn-S(n-1)=tn²+(t-9)n+t-3/2-t(n-1)²-(t-1)(n-1)-t+3/2
=t[n²-(n-1)²]+(t-9)[n-(n-1)]=t(2n-1)+(t-9)=2tn-9
a(n+1)-an=2t(n+1)-9-2tn+9=2t,即从第3项起后一项与前一项的差为常数2t
要使成为等差数列,只要使a2-a1=2t
a2-a1=4t-9-(3t-21/2)=t+3/2=2t
∴t=3/2
∴a1=3t-21/2=9/2-21/2=-6,an=2tn-9=3n-9(n=1时也符合)
∴an通项为an=3n-9

Answer 2

先加点金币落，我正缺钱花