请大侠帮忙.数字1.0.3.4.13.12.39…的通项公式
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/30 04:33:02
请大侠帮忙.数字1.0.3.4.13.12.39…的通项公式
如题.可以的话请写出推导.
如题.可以的话请写出推导.
![请大侠帮忙.数字1.0.3.4.13.12.39…的通项公式](/uploads/image/z/15744257-17-7.jpg?t=%E8%AF%B7%E5%A4%A7%E4%BE%A0%E5%B8%AE%E5%BF%99.%E6%95%B0%E5%AD%971.0.3.4.13.12.39%E2%80%A6%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
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.