To Prove:
4n + 15n – 1 is divisible by 9.
Let us prove this question by principle of mathematical induction (PMI) for all natural numbers
Let P(n):4n + 15n – 1 is divisible by 9.
For n = 1 P(n) is true since 4n + 15n – 1 = 41 + 15 x 1 – 1 = 18
Which is divisible of 9
Assume P(k) is true for some positive integer k , ie,
= 4k + 15k – 1 is divisible by 9
∴ 4k + 15k – 1 = m x 9, where m ∈ N …(1)
We will now prove that P(k + 1) is true whenever P( k ) is true.
Consider,
= 4k+1 + 15(k + 1) – 1
= 4k x 4 + 15k + 15 - 1
= 4k x 4 + 15k + 14 + (60k + 4) - (60k + 4)
[Adding and subtracting 60k + 4]
= 9 x r , where r = [(4m) - (5k - 2)] is a natural number
Therefore 4k + 15k - 1 is a divisible of 9
Therefore, P (k + 1) is true whenever P(k) is true
By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N.
Hence proved.