当n属于N时,Sn=1-(1/2)+(1/3)-(1/4)+……+(1/2n-1)-(1/2n).Tn=(1/n)+(1/n+2)+(1/n+3)…+1/2n
问题描述：
当n属于N时,Sn=1-(1/2)+(1/3)-(1/4)+……+(1/2n-1)-(1/2n).Tn=(1/n)+(1/n+2)+(1/n+3)…+1/2n
(1)求S1,S2,T1,T2
(2)猜想Sn与Tn的关系,并用数学归纳法证明
答
(1)S1=1-1/2=1/2,S2=1-1/2+1/3-1/4=7/12
T1=1+1/2=3/2,T2=1+1/3+1/4=19/12
(2)Tn-Sn=1
T1-S1=1,假设Tn-Sn=1,
T(n+1)=Tn + 1/(2n+1) + 1/[2(n+1)]
S(n+1)=Sn + 1/[2(n+1)-1] + 1/[2(n+1)]= Sn + 1/(2n-1) + 1/[2(n+1)]
则T(n+1)-S(n+1)= Tn - Sn = 1
所以原假设得证