A. Maths ( F.4 ) 急急急

Let 6^(k+1) - 5(k+1) - 1 = 25m where m is an integer.
6^(k+1) = 5(k+1) +1+ 25m
for n=k+1
LHS
=6^(k+1+1) - 5(k+1+1) - 1
=6^(k+1) * 6 - 5(k+2) - 1
=[5(k+1) +1+ 25m] * 6 - 5k-10-1
=30(k) +30 + 6 + 150m - 5k- 10 -1
=25(k) +150m + 25
... divisible by 25!

For n=1

6^2-5*2-1=25 which is divisible by 25

Assume n=k is true

let n=k+1

6^(k+2)-5(k+2)-1
= 6*6^(k+1) - 5(k+1) -5 -1
= 6*( 6^(k+1) -5(k+1) -1) + 5*5(k+1) -5-1+6
= 6*25m+25(k+1) where m is positive integer
=25( 6m+k+1) which is divisible by 25

so n=k+1 is true

By MI, 6^(n+1) - 5(n+1) -1 is divisible by 25, for all positive integers n

2007-01-10 19:06:36 補充：
係 Assume n=k is true 後面加番 where k is some positive integer