已知数列an满足a1=1,前n项的和为Sn 且对任意的n∈N*有(n+1)an-2Sn=3n-3成立1.求a2 ,a3的值并推导_数学解答

已知数列an满足a1=1,前n项的和为Sn 且对任意的n∈N*有(n+1)an-2Sn=3n-3成立
1.求a2 ,a3的值并推导an的通项公式
2.记数列1/an的前n项和为Tn,若T2n+1-Tn小于等于m/15 对n∈N*恒成立,试确定正整数m的最小值（注：n、n+常数都为下标）

人气：469 ℃　时间：2019-09-27 18:53:41

解答

(1)
(n+1)an-2Sn=3n-3
2Sn = (n+1)an -3n+3
an = Sn -S(n-1)
2an = (n+1)an -na(n-1) -3
(n-1)an =na(n-1) +3
an/n - a(n-1)/(n-1) = 3/[n(n-1)]
= 3[ 1/(n-1) - 1/n]
an/n - a1/1 = 3[1- 1/n]
an/n = (4n- 3)/n
an = 4n-3
a2= 5
a3= 9
(2)
bn =1/an
Tn = b1+b2+...+bn
T(2n+1)-Tn
= 1/(4n+1) +1/(4n+5)+...+1/(8n+1)
max (T(2n+1) -Tn) at n=1
T3 -T1
= 1/5+1/9
= 14/45