P(n) = n³ + 5n
P(1) = 1 + 5
P(1) = 6
We assume P(k) is true and divisible by 6.
P(k) = k³ + 5k is divisible by 6 and can be written as 6c or 3 x 2c
We need to prove that P(k + 1) is divisible by 6
P(k + 1) = (k + 1)³ + 5(k + 1)
P(k + 1) = k³ + 1 + 3k² + 3k + 5k + 5
P(k + 1) = (k³ + 5k) + 3k² + 3k + 6
P(k + 1) = 6c + 3(k² + k + 2)
P(k + 1) = (3 x 2c) + 3(k² + k + 2)
Therefore, P(k + 1) is definitely divisible by 3

P(n) = 10³ⁿ + 2^{4k + 1}. 9 + k
P(1) = 10³ + 2⁵ . 9 + k
P(1) = 1000 + 288 + k
P(1) = 1288 + k
When 1288 is divided by 11, remainder is 1.
Therefore, 1287 is divisible by 11.
The next number that is divisible is 1298.
k = 1298 ? 1288
k = 10

P(n) = n³ (n + 1)
P(k + 1) = (k + 1)³ (k + 1 + 1)
P(k + 1) = (k³ + 3k² + 3k + 1) (k + 2)
P(k + 1) = k⁴ + 3k³ + 3k² + k + 2k³ + 6k² + 6k + 2
P(k + 1) = k⁴ + 5k³ + 9k² + 7k + 2

In mamatical induction, if P(k) is true, we need to prove that P(k + 1) is also true. Here P(k + 1) is found by substituting (k + 1) in place of k. P(k + 1) = (k + 1)² (k + 1 + 3) ((k + 1)² ? 1)
P(k + 1) = (k + 1)² (k + 4) (k² + 1 + 2k ? 1)
P(k + 1) = (k + 1)² (k + 4) (k² + 2k)
P(k + 1) = (k + 1)² (k + 4) k (k +2)

P(n) = n(n² ? 1)
P(k + 1) = (k + 1) ((k + 1)² ? 1)
P(k + 1) = (k + 1) (k² + 1 + 2k ? 1)
P(k + 1) = (k + 1) (k² + 2k)
P(k + 1) = (k + 1) k (k + 2)
Therefore, k, (k + 1), (k ? 1) divide P(k + 1).

P(n) = n² + 3n
P(1) = 1 + 3
P(1) = 4
Let?s assume that P(k) is true and divisible by 4. Therefore, P(k) = k² + 3k can be written as 4c.
We need to check if P(k + 1) is divisible by 4
P(k+1) = (k + 1)² + 3(k + 1)
P(k+1) = k² + 1 + 2k + 3k + 3
P(k+1) = k² + 5k + 4
P(k+1) = (k² + 3k) + 2k + 4
P(k+1) = 4c + 2k + 4
P(k+1) = 4c + 2(k + 2)
Clearly second part of equation is not divisible by 4. However P(k) = 4c is divisible by 2 and
P(k + 1) is also divisible by 2. Therefore, 2 divides P(n).

When we use principle of mamatical induction, we assume that P(n) is true for P(k) and prove that P(k + 1) is also true. Here P(k) is k(k + 1)! < k!

P(n) = n(n + 1)/2
P(1) = 1
We assume P(k) to be true, refore, P(k) = \(\frac{k(k + 1)}{2}\)
To prove that, P(k + 1) = \(\frac{(k+1)(k+ 2)}{2}\)
Proof:
P(k + 1) = 1 + 2 + 3 +?.+ k + k + 1
P(k + 1) = \(\frac{k(k + 1)}{2}\) + k+1
P(k + 1) = \(\frac{k(k+ 1)+2(k+1)}{2}\)
P(k + 1) = \(\frac{(k+1)(k+ 2)}{2}\)
Therefore, P(n) is true by principle of mamatical induction.

P(n) = 7²ⁿ + 2^{2n ? 2} . 3^{n ? 1}
P(1) = 7² + 2⁰ . 3⁰
P(1) = 50
We now assume that P(k) is divisible by 50.
Therefore, P(k) = 7^2k + 2^{2k ? 2} . 3^{k ? 1}
To prove P(k + 1) = 7^{2(k ? 1)} + 2^{2(k + 1) – 2} . 3^{(k + 1) ? 1} is divisible by 50
P(k + 1) = 7^2k . 7² + 2^2k . 3^k
P(k + 1) = 7² ( 7^2k + 2^{2n ? 2} . 3^{k ? 1} ? 2^{2k ? 2} . 3^{k ? 1} ) + 2^2k . 3^k
P(k + 1) = 7² x 50c ? 7² . 2^{2k ? 2} . 3^{k ? 1} + 2^2k . 3^k
Since P(k) = 7^2k + 2^{2n ? 2} . 3^{k ? 1} ? 2^{2k ? 2} . 3^{k ? 1} is divisible by 50, it can be written as 50c.
P(k + 1) = 49 x 50C – 49 . 2^{2k ? 2} . 3^{k ? 1} + 3 x 4 x 2^{2k ? 2} . 3^{k ? 1}
P(k + 1) = 49 x 50C – 2^{2k ? 2} . 3^{k ? 1} x 37
While first term is divisible by 50, second term is not.
Therefore, by principle of mamatical induction, 7²ⁿ + 2^{2n ? 2} . 3^{n ? 1} is not divisible by 50.

P(n) = 2^{4n ? 1}
P(1) = 2³ = 8
Let us assume P(k) is divisible by 8 and can be written as 8c, where c is any integer.
P(k) = 2^{4k – 1} = 8c
P(k + 1) = 2^{4(k + 1) ? 1}
P(k + 1) = 2^{4k + 3}
P(k + 1) = 2⁴ . 2^{4k ? 1}
P(k + 1) = 2⁴ . 8c
Clearly, P(k + 1) is divisible by 2, 4, 8 and 16.