已知数列{a n }与{b n }满足b n+1 a n +b n a n+1 =(-2) n +1,b n = 3+( -1) n-1 2 ,
已知数列{an}与{bn}满足bn+1an+bnan+1=(-2)n+1,bn=3+(-1)n-12,n∈N*,且a1=2.(Ⅰ)求a2,a3的值(Ⅱ)设cn=a2n+1...
已知数列{a n }与{b n }满足b n+1 a n +b n a n+1 =(-2) n +1,b n = 3+( -1) n-1 2 ,n∈N * ,且a 1 =2.(Ⅰ)求a 2 ,a 3 的值(Ⅱ)设c n =a 2n+1 -a 2n-1 ,n∈N * ,证明{c n }是等比数列(Ⅲ)设S n 为{a n }的前n项和,证明 S 1 a 1 + S 2 a 2 +…+ S 2n-1 a 2n-1 + S 2n a 2n ≤n- 1 3 (n∈N * )
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(Ⅰ)由b n =
又b n+1 a n +b n a n+1 =(-2) n +1, 当n=1时,a 1 +2a 2 =-1,可得由a 1 =2,a 2 =-
当n=2时,2a 2 +a 3 =5可得a 3 =8; (Ⅱ)证明:对任意n∈N * , a 2n-1 +2a 2n =-2 2n-1 +1…① 2a 2n +a 2n+1 =2 2n +1…② ②-①,得a 2n+1 -a 2n-1 =3×2 2n-1 ,即:c n =3×2 2n-1 ,于是
所以{c n }是等比数列. (Ⅲ)证明: a1=2,由(Ⅱ)知,当k∈N * 且k≥2时, a 2k-1 =a 1 +(a 3 -a 1 )+(a 5 -a 3 )+(a 7 -a 5 )+…+(a 2k-1 -a 2k-3 ) =2+3(2+2 3 +2 5 +…+2 2k-3 )=2+3×
故对任意的k∈N * ,a 2k-1 =2 2k-1 . 由①得2 2k-1 +2a 2k =-2 2k-1 +1,所以 a 2k =
因此, S 2k =( a 1 + a 2 )+( a 3 + a 4 )+…+( a 2k-1 + a 2k ) =
于是, S 2k-1 = S 2k - a 2k =
故
= 1-
所以,对任意的n∈N * ,
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