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Using the principle of mathematical induction, prove each of the following for all n ϵ N: 

(4n + 15n – 1) is divisible by 9.

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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.

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