Question

已知等差数列{An}的前n项和为Sn=pn^2-2n+q,(p,q属于R,n属于N)

Answer 1

1)A1=S1=p-2+q,

An=Sn-Sn-1=pn^2-2n+q-p(n-1)^2+2(n-1)-q=p(2n-1)-2=2pn-p-2

d=An-An-1=2p

An=A1+(n-1)*d=p-2+q+2(n-1)p=2pn-p-2+q = 2pn-p-2

q=0

Sn=pn^2-2n,An=2pn-p-2,d=2p,A1=p-2

2) A3=A1+2d=p-2+4p=5p-2=18,p=4

A1=2,An=8n-6,

8n-6=2log2Bn,log2Bn=4n-3,Bn=2^(4n-3),Bn/Bn-1=2^4=16,Bn是公比为16的等比数列,首项B1＝2,设Bn的前n项和是Tn

Tn=2*(16^n-1)/(16-1)=2*(16^n-1)/15