✔ 最佳答案
For n=1,
(3+1)7-1=9,which is divisible by 9.
∴The statement is true for n=1.
Let (3k+1)7k-1= 9A ,where k and A are positive integers.
For n=k+1,
=[3(k+1)+1]7k+1-1
=(3k+4)7k+1-[(3k+1)7k -9A ]
=7k[ 7(3k+4)- (3k+1)]+ 9A
=7k(18k-27)+ 9A
=7k[9(2k-3)+ 9A ]
=9[7k(2k-3)+ A]
∴[3(k+1)+1]7k+1-1is also divisible by 9.
∴The statement is also true for n=k+1 if it is true for n=k.
By the principle of mathematical induction,the statement is true for all positive integer n.
2008-09-11 17:51:08 補充：
For n=1,
(3+1)7-1=27,which is divisible by 9.
∴The statement is true for n=1.
2008-09-11 17:51:25 補充：
以上才正確
