已知数列{an}的前N项和为Sn 且an+1=Sn-n+3,a1=2,设Bn=n/Sn-n+2前N项和为Tn 求证Tn 小于4/3_数学解答

已知数列{an}的前N项和为Sn 且an+1=Sn-n+3,a1=2,设Bn=n/Sn-n+2前N项和为Tn 求证Tn 小于4/3

人气：306 ℃　时间：2019-10-17 07:27:31

解答

Sn+1 —Sn=an+1=Sn—n+3,即Sn+1=2Sn-n+3,所以Sn+1 -(n+1)+2=2(Sn-n+2)又S1 -1+2=3,所以Sn-n+2=3*2^n-1,所以bn=n/(3*2^n-1),Tn=1/3 (1/2^0 +2/2 ^1+3/2^2+4/2^3+.+n/2^n-1) .10.5Tn=1/3 ( 1/2^1+2/2^2+.+n-1/2^n-1 +n...