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设其为 A_n
由题有：
A_4 - A_2 = 4; A_6 -A_4 = 8; ==>
A_2*n+2 -A_2*n = 4*2^(n-1) ===>
A_2*n = 4*(2^n -1); n = 1,2,3.
A_3 - A_1 = 3 - 1 = 2 = 1+1 = 1^2 +1;
A_5 - A_3 = 13 - 3 = 10 = 9+1 = 3^2 +1;
A_7 - A_5 = 39 - 13 = 26 = 25+1 = 5^2 +1; ==>
A_2*n+1 - A_2*n-1 = (2*n - 1)^2 +1; ==>
A_2n+1 = n + 1 + n*(4*n^2-1)/3; n = 0,1,2.
综上知A_n为：
当 n 为偶数时.
当 n 为奇数时.
由题有：
A_4 - A_2 = 4; A_6 -A_4 = 8; ==>
A_2*n+2 -A_2*n = 4*2^(n-1) ===>
A_2*n = 4*(2^n -1); n = 1,2,3.
A_3 - A_1 = 3 - 1 = 2 = 1+1 = 1^2 +1;
A_5 - A_3 = 13 - 3 = 10 = 9+1 = 3^2 +1;
A_7 - A_5 = 39 - 13 = 26 = 25+1 = 5^2 +1; ==>
A_2*n+1 - A_2*n-1 = (2*n - 1)^2 +1; ==>
A_2n+1 = n + 1 + n*(4*n^2-1)/3; n = 0,1,2.
综上知A_n为：
当 n 为偶数时.
当 n 为奇数时.
1年前
10
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