F.4 maths

Let P(n) be the proposition
" 6^n x(5n-1)+1 is divisible by 25 "
When n=1
6^1 x(5*1-1)+1
=6*4+1
=25
is divisible by 25
So P(1) is true.
Assume that P(k) is true for some positive integers n
i.e. 6^k (5k-1)+1=25M, where M is an integer
When n=k+1
6^k+1 [5(k+1)-1]+1= 6^k+1 (5k+5-1)+1
= 6^k*6(5k+4)+1
= 6^k*6(5k+4)+1+25M-6^k (5k-1)-1
= 6^k [6(5k+4)-(5k-1)]+25M
= 6^k (30k+24-5k+1)+25M
= 6^k (25k+25)+25M
= 25(6^k k +6^k+M)
= 25[6^k (k+1)+M]
So P(k+1) is true
By the Mathematical induction, 6^n x(5n-1)+1is divisible by 25 for all positive integers n

2006-11-19 18:45:52 補充：
New Progress in Additional Mathematics 1 有le題

2006-11-19 18:47:59 補充：
Revision Ex 3 (21)

Prove , by mathematical induction, that 6^n x(5n-1)+1is divisible by 25 for all positive integers n
(1)When n = 1
6^n(5n-1)+1 = 6*4+1 =25à是25的倍數，所以n=1時成立。
(2)設 n = k 是成立，即6^k(5k-1)+1是 25的倍數
(3)當 n = k + 1
LHS
= 6^(k+1)(5(k+1)-1)+1
=6^k*6*(5k+5-1)+1
=6^k*(30k+30-6)+1
=6^k(30k+24)+1
=6^k(5k-1+25k+25) + 1
=6^k (5k-1) + 6^k (25k+25) + 1
=6^k (5k-1) +1 + 6^k *25(k+1)
As 6^k (5k-1) +1 is assumed to be divisible by 25 and 6^k*25(k+1) has a mulitplier of 25, so 6^n(5n-1)+1 is divisible by 25 for all positive integers n