f.4 amaths

when n=1, =5^0 - 3^0 - 2^0=0
when n=2, =5^3 - 3^3 - 2^3=90,which is divisible by 15
assume n=k
i.e. [5^(2k-1)]-[3^(2k-1)]-[2^(2k-1)]=15M,where M is an integer.
when n=k+1
={5^[2(k+1)-1]} - {3^[2(k+1)-1]} - {2^[2(k+1)-1]}
=5^(2k+2-1) - 3^(2k+2-1) - 2^(2k+2-1)
=[5^(2k-1)][25] - [3^(2k-1)][9] - [2^(2k-1)][4]
=[5^(2k-1)][21+4] - [3^(2k-1)][5+4] - [2^(2k-1)][4]
take out the common factor
= {[5^(2k-1)][21] + [5^(2k-1)][4]} - {[3^(2k-1)][5}+3^(2k-1)][4]} - [2^(2k-1)][4]
= [5^(2k-1)][21] + [5^(2k-1)][4] - [3^(2k-1)][5] - 3^(2k-1)][4] - [2^(2k-1)][4]
= [4][5^(2k-1) - 3^(2k-1) - 2^(2k-1)] + [5^(2k-1)][21] - [3^(2k-1)][5]
=(4)(15M)+ [5^(2k-1)][21] - [3^(2k-1)][5]
=(4)(15M)+ {[(5^(2k-2)](5)(3)(7)} - [3^(2k-2)(3)][5]
=(4)(15M)+ {[(5^(2k-2)](15)(7)} - [3^(2k-2)(15)]
=15{4+[(5^(2k-2)](7) - [3^(2k-2)]}
By the principal of mathematical induction , [5^(2n-1)] -[3^(2n-1)]-[2^(2n-1)]is divisible by 15 for all positive integers n.

Prove that 5^(2n-1)-3^(2n-1)-2^(2n-1) is divisible by15 for all natural numbers n.
(1)When n = 1
5^(2-1)-3^(2-1)-2^(2-1) = 5-3-2 =0,0是15的倍數，所以n=1時成立。
(2)設 n = k 是成立，即5^(2k-1)-3^(2k-1)-2^(2k-1) 是 15的倍數
(3)當 n = k + 1
LHS
= 5^(2(k+1)-1)-3^(2(k+1)-1)-2^(2(k+1)-1)
=5^(2k+2-1)-3^(2k+2-1)-2^(2k+2-1)
=5^(2k-1)*5^2-3^(2k-1)*3^2-2^(2k-1)*2^2

=5^(2k-1)*25-3^(2k-1)*9-2^(2k-1)*4
And it can be expanded as:
=21*(5^(2k-1))+4(5^(2k-1))-5(3^(2k-1))-4(3^(2k-1))- 2^(2k-1)*4
=4〔5^(2k-1)-3^(2k-1) - 2^(2k-1)〕+21*(5^(2k-1)) -5(3^(2k-1))
=4〔5^(2k-1)-3^(2k-1) - 2^(2k-1)〕+3*7*(5^(2k-1)) -5(3^(2k-1))
First,
4〔5^(2k-1)-3^(2k-1)-2^(2k-1) 〕is assumed to be divisible by 15
Second,
In 3*7*(5^(2k-1)), 3*7*5 is divisble by 15, so this can be divided by 15
Third,
In 5(3^(2k-1)), 5*3 is equal to 15, so this one can also be divided by 15
So the equation 5^(2n-1)-3^(2n-1)-2^(2n-1) is divisible by15 for all natural numbers n.