cml008
春芽
共回答了12个问题采纳率:91.7% 举报
(I)设公差为d,由已知得:
S 4 =14
a 3 2 = a 1 a 7 ,
即
4 a 1 +
4×3
2 d=14
( a 1 +2d) 2 = a 1 ( a 1 +6d) ,
解得:d=1或d=0(舍去),
∴a 1 =2,
故a n =2+(n-1)=n+1;
(II)∵
1
a n a n+1 =
1
(n+1)(n+2) =
1
n+1 -
1
n+2 ,
∴T n =
1
2 -
1
3 +
1
3 -
1
4 +…+
1
n+1 -
1
n+2 =
1
2 -
1
n+2 =
n
2(n+2) ,
∵T n ≤λa n+1 对∀n∈N * 恒成立,即
n
2(n+2) ≤λ(n+2),λ≥
n
2 (n+2) 2 ∀n∈N * 恒成立,
又
n
2 (n+2) 2 =
1
2(n+
4
n +4) ≤
1
2(4+4) =
1
16 ,
∴λ的最小值为
1
16 .
1年前
2