a1 a3 1, (a3 a5) 1,a5 a7 1,
( 1)k(a2k 3 a2k 1) 1.
k
a ( 1)a2k 1 (k 1), 1将以上各式相加,得k 1
a ( 1)(k 1), 2k 1即
k 1
a ( 1)(k 3). 2k此式当k=1时也成立.由④式得
从而
S2k (a2 a4) (a6 a8) (a4k 2 a4k) k,
S2k 1 S2k a4k k 3.
*
n N,n 2, 所以,对任意
n
SkS4m 3S4m 2S4m 1S4m
( ) aaaaak 1km 14m 34m 24m 14m
4n
(
m 1n
n
2m 22m 12m 32m
)2m2m 22m 12m 3 23
)
2m(2m 1)(2m 2)(2m 2)
(
m 1
n
253
2 3m 22m(2m 1)(2n 2)(2n 3)
1n53
3m 2(2m 1)(2m 1)(2n 2)(2n 3)
151111113 [( ) ( ) ( )] 3235572n 12n 1(2n 2)(2n 3)
15513
3622n 1(2n 2)(2n 3)7 .6
对于n=1,不等式显然成立.
*
n N, 所以,对任意