二項次及換底對數

原式即 log(x+1) / log3 + log(x+1) / log5 = 1.5log5 * log(x+1) + log3 * log(x+1) = 1.5 log3 log5log(x+1) * (log5 + log3) = 1.5 log3 log5log(x+1) = (1.5 log3 log5) / (log3 + log5)log(x+1) = 0.42534x+1 = 2.66282x = 1.66282
1)(x + 1)^99= x^99 + (99C1)x^98 + (99C2)x^97 + ... + (99C97)x^2 + (99C98)x + 1= [x^99 + (99C1)x^98 + (99C2)x^97 + ... + (99C97)x^2] + 99x + 1= (x^2) f(x) + 99x + 1 ,
f(x) = x^97 + (99C1)x^96 + (99C2)x^95 + ... (99C96)x + (99C97)以6代入x :7^99= (6 + 1)^99= (6^2) f(6) + 99(6) + 1= 36 f(6) + 99(6) + 1= 9 [4f(6) + 11(6)] + 1餘數 = 1
2)13^90= (12 + 1)^90= 12^90 + (90C1)12^89 + (90C2)12^88 + (90C3)12^87 + ... + (90C88)12^2
+ (90C89)12 + 1= 36K + (90C89)12 + 1= 36K + 13餘數 = 13
3)13^n - 4^n= (18 - 5)^n - (9 - 5)^n=18^n - (nC1)(18^(n-1))(5) + (nC2)(18^(n-2))(5^2) -...+(nC(n-1))18(5^(n-1)) - 5^n- [9^n - (nC1)(9^(n-1))(5) + (nC2)(9^(n-2))(5^2) - ... + (nC(n-1))9(5^(n-1)) - 5^n]= (18^n - 9^n) - 5(nC1)(18^(n-1) - 9^(n-1)) + (5^2)(nC2)(18^(n-2) - 9^(n-2)) - ...
+ (5^(n-1))(nC(n-1))(18 - 9)上式各項均被 9 整除 。

2010-09-16 23:13:31 補充：
更正2) :

= 36K + (90C89)12 + 1

= 36K + 90*12 + 1

= 36K + 30 * 36 + 1

= 36(K + 30) + 1

餘數 = 1

2010-09-16 23:13:40 補充：
更正2) :

= 36K + (90C89)12 + 1

= 36K + 90*12 + 1

= 36K + 30 * 36 + 1

= 36(K + 30) + 1

餘數 = 1

2010-09-17 11:38:13 補充：
3)較好方法 :

13^n - 4^n

= (9 + 4)^n - 4^n

= 9^n + (nC1)(9^(n-1))4 + ... + 4^n - 4^n

= 9K + 4^n - 4^n

= 9K 被9整除

2010-09-17 11:38:28 補充：
3)較好方法 :

13^n - 4^n

= (9 + 4)^n - 4^n

= 9^n + (nC1)(9^(n-1))4 + ... + 4^n - 4^n

= 9K + 4^n - 4^n

= 9K 被9整除

2010-09-20 16:35:00 補充：
36K = 12^90 + (90C1)12^89 + (90C2)12^88 + (90C3)12^87 + ... + (90C88)12^2

12^90 + (90C1)12^89 + (90C2)12^88 + (90C3)12^87 + ... + (90C88)12^2

是36的倍數。因為每項含 12^2 = 4 * 36 。

我想問第二題, 36k是什麼??