因为S2n=2n(a1+a2n)/2=n[2a1+(2n-1)d] ,Sn=n(a1+an)/2=n[2a1+(n-1)d]/2又S2n/Sn=4n+2/n+1,所以[2+(2n-1)d]/[2+(n-1)d]=(2n+1)/(n+1)对任意正整数n都成立,解得d=1,于是An=n,Bn=np^n,(1)当p=1时,Bn前n项和为Tn=n(n+1)/...