F4 M2 (mathematical induction)

(1a) Let P(n) be the proposition that 7^(3n - 1) + 8 is divisible by 19 for all positive integers n
P(1) is true since 7^2 + 8 = 57 = 19*3
Assume P(k) is true, i.e. 7^(3k - 1) + 8 = 19N where N is an integer
=> 7^(3k - 1) = 19N - 8
then 7^(3k + 3 - 1) + 8
= (7^3)(7^3k - 1) + 8
= 343(19N - 8) + 8
= (343)(19N) - 2736
= (343)(19N) - 144 * 19
= 19(343N - 144) is divisible by 19 => P()k + 1) is true
Hence by the principle of MI, P(n) is true for alll positive integers n
(b) 7^2009 + 21 = 7^(3*670 - 1) + 8 + 13
There is the remainder is 13
(2a)(i) Let P(n) be the proposition that 5^n - 1 is divisible by 4 for all positive integers n
P(1) is true since 5^1 - 1 = 4
Assume P(k) is true, i.e. 5^k - 1 = 4N where N is an integer
then 5^(k + 1) - 1
= 5*5^k - 1
= 5(4N + 1) - 1
= 20N - 4
= 4(5N - 1) is divisible by 4 => P()k + 1) is true
Hence by the principle of MI, P(n) is true for alll positive integers n
(ii) Let P(n) be the proposition that 4^n - 1 is divisible by 3 for all positive integers n
P(1) is true since 4^1 - 1 = 3
Assume P(k) is true, i.e. 4^k - 1 = 3N where N is an integer
then 4^(k + 1) - 1
= 4*4^k - 1
= 4(3N + 1) - 1
= 12N - 3
= 3(4N - 1) is divisible by 3 => P()k + 1) is true
Hence by the principle of MI, P(n) is true for alll positive integers n
(b) Let P(n) be the proposition that 3(5^n) + 4^(n + 1) - 7 is divisible by 12 for all positive integers n
P(1) is true since 3(5) + 4^2 - 7 = 24 is divisible by 12
Assume P(k) is true, i.e. 3(5^k) + 4^(k + 1) - 7 = 12N where N is an integer
then 3(5^k+1) + 4^(k + 2) - 7
= 15(5^k) + 4(4^k+1) - 7
= 5[3(5^k) + 4^(k+1) - 7] - 4^(k + 1) + 28
= 5(12N) - (4)(4^k) + 28
From (a) 4^k - 1 = 3M
4^k = 3M + 1
So 5(12N) - (4)(4^k) + 28 = 5(12N) - 4(3M + 1) + 28
= 5(12N) - 12M + 24
= 12(5N - M + 2) is divisible by 12 => P()k + 1) is true
Hence by the principle of MI, P(n) is true for alll positive integers n