F4 M.I please !!

Let P(n) be the proposition that 4(6^n)+5^(n+1)-9 is divisible by 20 for all positive integers n
For n = 1, 4*6 + 25 - 9 = 40 is divisible by 20 so P(1) is true
For n = 2, 144 + 125 - 9 = 260 is divisible by 20 so P(2) is true
Assuming that P(k) and P(k + 1) are true, i.e.
4(6^k) + 5^(k+1) - 9 = 20N where N is an integer ... (1)
4[6^(k+1)] + 5^(k+2) - 9 = 20M where M is an integer ... (2)
when n = k+2
4[6^(k+2)] + 5^(k+3) - 9
= (6)(4)[6^(k+1)] + (5)5^(k+2) - 9
= (5)(4)[6^(k+1)] + (5)5^(k+2) - 9 + (4)[6^(k+1)]
= 5{4[6^(k+1)] + (5)^(k+2) - 9} + 36 + (4)[6^(k+1)]
= 5(20M) + 6[6 + (4)6^k] from (2)
= 100M + 6[6 + 20N + 9 - 5^(k+1)] from (1)
= 100M + 6[15 + 20N - 5^(k+1)]
= 100M + 30(3 + 4N - 5^k)
3 + 4N - 5^k is odd + even - odd = even
so we can express 3 + 4N - 5^k as 2P where P is an integer
100M + 30(3 + 4N - 5^k) = 100M + 30(2P)
= 20(5M + 3P) is divisible by 20
Hence P(k+2) is also true
By MI, P(n) is true for all positive integers n.