已知数列{a n }满足 a 1 =1, a 2 = 1 2 ,且[3+(-1) n ]a n+2 -2a n +2[(-1) n -1
已知数列{an}满足a1=1,a2=12,且[3+(-1)n]an+2-2an+2[(-1)n-1]=0,n∈N*.(1)求a3,a4,a5,a6的值及数列{an}的通项...
已知数列{a n }满足 a 1 =1, a 2 = 1 2 ,且[3+(-1) n ]a n+2 -2a n +2[(-1) n -1]=0,n∈N*.(1)求a 3 ,a 4 ,a 5 ,a 6 的值及数列{a n }的通项公式;(2)设b n =a 2n-1 ?a 2n (n∈N * ),求数列{b n }的前n项和S n .
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(1) a 3 =3, a 4 = , a 5 =5, a 6 = 当n为奇数时,a n+2 =a n +2 所以a 2n-1 =2n-1(3分) 当n为偶数时, a n+2 = a n 即 a 2n = a 2 ?( ) n-1 =( ) n (5分) 因此,数列a n 的通项公式为 a n = (6分) (2)因为 b n =(2n-1)?( ) n S n =1? +3?( ) 2 +5?( ) 3 ++(2n-3)?( ) n-1 +(2n-1)?( ) n S n =1?( ) 2 +3?( ) 3 +5?( ) 4 ++(2n-3)?( ) n +(2n-1)?( ) n+1 两式相减得 S n =1? +2[( ) 2 ++( ) n ]-(2n-1)?( ) n-1 (8分) = + -(2n-1)?( ) n+1 = -(2n+3)( ) n+1 ∴ S n =3-(2n+3)?( ) n (12分) |
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