標題:
此文章來自奇摩知識+如有不便請留言告知
a math MI
發問:
1)1^3-2^3+3^3-4^3+...+(2n-1)^3 -(2n)^3= -n^2(4n+3) 2)41^n is divisable by 20 3)11^n+2 +12^2n+1 is divisable by 133 4)9 is a factor of 10^n+1 +3(10^n) +5
最佳解答:
1) 1^3 - 2^3 + 3^3 - 4^3 + ... + (2n-1)^3 - (2n)^3 = -n^2 (4n+3) -(1^2) (4(1)+3) = -7 = 1^3 - 2^3 assume 1^3 - 2^3 + 3^3 - 4^3 + ... + (2k-1)^3 - (2k)^3 = -k^2 (4k+3) consider 1^3 - 2^3 + ... + (2k-1)^3 - (2k)^3 + (2k+1)^3 - (2k+2)^3 = -k^2 (4k+3) + (2k+1)^3 - (2k+2)^3 = -(4k^3 + 15k^2 + 18k + 7) = -(k+1)^2 (4k+7) = -(k+1)^2 (4(k+1)+3) so, by M.I., 1^3 - 2^3 + 3^3 - 4^3 + ... + (2n-1)^3 - (2n)^3 = -n^2 (4n+3) ------------------------------------------------------------- 2) 41^n is divisable by 20 有無打錯野呀??? 41^1 = 41 is not divisable by 20 ------------------------------------------------------------- 3) 11^n+2 +12^2n+1 is divisable by 133 11^(1+2) + 12^(2(1)+1) = 3059 is divisable by 133 assume 11^(k+2) + 12^(2k+1) is divisable by 133 so, 11^(k+2) + 12^(2k+1) = 133m, m is integer 11^(k+1+2) + 12^(2(k+1)+1) = 11 [11^(k+2) ] + 144 [12^(2k+1)] = 11 [11^(k+2) ] + 144 [133m - 11^(k+2)] = 133[144m - 11^(k+2)] is divisable by 133 because 144m - 11^(k+2) is integer ------------------------------------------------------------- 4) 9 is a factor of 10^n+1 +3(10^n) +5 10^(1 + 1) + 3(10^1) + 5 = 135 divisable by 9 assume 9 is a factor of 10^(k+1) + 3(10^k) + 5 so, 10^(k+1) + 3(10^k) + 5 = 9m, m is integer consider 10^(k+2) + 3[10^(k+1)] + 5 = 10[10^(k+1)] + 30(10^k) + 5 = 10[9m - 3(10^k) - 5] + 30(10^k) + 5 = 90m - 45 = 9(10m - 5) divisable by 9 because 10m - 5 is integer so, by M.I., 9 is a factor of 10^(n+1) + 3(10^n) + 5 -------------------------------------------------------------
其他解答:
留言列表
