Question

1.等差数列an前n项和为Sn,求证S（2n-1）=（2n-1）an

Answer 1

1.等差数列an前n项和为Sn,设首项为a1公差为d,an=a1+(n-1)d；a(2n-1)=a1+(2n-1-1)d=a1+2(n-1)d,S(2n-1)=[a1+a1+2(n-1)d](2n-1)/2=[a1(n-1)d](2n-1)=(2n-1)an；

2.Sn/Tn=2n/3n+1,S(2n-1)/T(2n-1)=[(2n-1)an]/[(2n-1)bn]=(4n-2)/[6n-3+1]=(2n-1)/(3n-1),

an/bn=(2n-1)/(3n-1).