2條M.I.

1.let P(n) be the proposition of (5^n)(4n-1)+1 is divisible by 16
When n=1, (5)(4-1)+1 = 16 is divisible by 16
∴P(1) is ture.
Assume P(k) is ture for any positive integers.
i.e. (5^k)(4k-1)+1=16M,where M is an integer.
then,[5^(k+1)] [4(k+1)-1]+1
=(5^k．5) (4k+3 ) +1
=[(16M-1)÷(4k-1)．5(4k+3)] +1
=[(16M-1)÷(4k-1)．(20k+15)] +1
=(320M-240M-20k-15)÷(4k-1) +1
=(320Mk-240M-20k-15+4k-1)÷(4k-1)
=(320Mk-240k-16k-16)÷(4k-1)
=16[(20Mk-15k-k-1)÷(4k-1)]
=16N,where N=(20Mk-15k-k-1)÷(4k-1) is an integer.
∴P(k+1) is ture.
By the principle of M.I. , P(n)is ture for all positive integers.

2.let P(n) be the proposition of n³+5n is divisible by 6
when n=1, 1+5 = 6 is divisible by 6
∴P(1) is ture.
Assume P(k) is ture for any natural numbers .
i.e. k³+5k=6M,whereM is an integer
then,(k+1)³+5(k+1)
=(k³+3k^2+3k+1) + 5k+5
=(6M-5k+3k^2+3k+1) + 5k+5
=6M+3k^2+3k+6
=6M+3k(k+1)+6
Since k and k+1 are continuous,either odd and even or even and odd.
Therefore,k(k+1)is a even number.3k(k+1)is divisible by 6
∴P(k+1) is ture.
By the principle of M.I.,P(n)is ture for all natural numbers .

1.

Assume n = k is true for some positive integer k.
i.e. 5k(4k - 1) + 1 = 16N, N is a integer.
For n = k + 1,
5k+1[4(k + 1) - 1] + 1
= 5k+1[(4k - 1) + 4] + 1
= 5k+1(4k - 1) + 1 + 4(5k+1)
= 5(5k)(4k - 1) + 5 + 4(5k+1) - 4
= 5[(5k)(4k - 1) + 1] + 4[(5k+1) - 1]
= 5(16N) + 4[5(5k) - 1]
= 16(5N) + 20(5k) - 4
= 16(5N) + 16(5k) + 4(5k) - 4
= 16(5N) + 16(5k) + 4[(5k) - 1]
= 16(5N) + 16(5k) + 4[(4 + 1)k - 1]
= 16(5N) + 16(5k) + 4[(4k + kC14k-1 + kC24k-2 + ... + kCk-141 + 1) - 1]
= 16(5N) + 16(5k) + 4[4(4k-1 + kC14k-2 + kC24k-3 + ... + kCk-1)]
= 16(5N) + 16(5k) + 16(4k-1 + kC14k-2 + kC24k-3 + ... + kCk-1)
= 16[(5N) + (5k) + (4k-1 + kC14k-2 + kC24k-3 + ... + kCk-1)]

2.

Assume n = k is true for some natural number k.
i.e. k3 + 5k = 6N, N is a integer.
For n = k + 1,
(k + 1)3 + 5(k + 1)
= (k3 + 3k2 + 3k + 1) + (5k + 5)
= k3 + 5k + 3k2 + 3k + 6
= 6N + 3k(k + 1) + 6
Since k and (k + 1) are respectively, either odd and even or even and odd, k(k + 1) is an even number and thus 3k(k + 1) is the multiple of 6.

2) Let P(n) be the statement.
For P(1), (1)^3+5(1)=6
which is divisible by 6
so, P(1) is true
Assume P(k) is true,
i.e. k^3+5k=6a, which a is a positive integer
For P(k+1), (k+1)^3+5(k) = k^3+3k^2+3k+1+5(k+1)
= (k^3+5k)+3k^2+3k+6
= 6a+3k^2+3k+6
= 6(a+1)+3k(k+1)
which 6(a+1) is divisible by 6
3k(k+1) is also divisible by 6. It is because either k or k+1 is even number
so, P(k+1) is true
By M.I. the statement is true for all natural number n.