# Proof and reasoning:- Use mathematical induction to prove that 4 x 6^(2n) + 3 x 2^(3n) is divisible by 7 for n=1,2,3,...........

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Let us assume 4 x 6^(2n) + 3 x 2^(3n) is divisible by 7

So for the next power (n+1):

4 x 6^[2(n+1)] + 3 x 2^[3(n+1)]

= 4 * 6^2n*6^2 + 3 * 2^3n*2^3

= 4 * 6^2n*(35 + 1) + 3 * 2^3n*(7 + 1)

= [4 x 6^(2n) + 3 x 2^(3n)] + [140*6^2n + 21*2^3n]

The first part (within third bracket) is divisible by 7, because we assumed it.

And the terms in the second part (within third bracket) is divisible by 7, because 140 and 21 is divisible by 7.

So if that assumption is correct for any particular value of n, it will be correct for all higher (integer) powers.

Plugging in n = 1, we can check that:

4 x 6^(2*1) + 3 x 2^(3*1)

= 4 x 6^(2) + 3 x 2^(3)

= (4 x 36) + (3 x 8)

= 168

This is divisible by 7 as 168/7 = 24

So we can conclude that 4 x 6^(2n) + 3 x 2^(3n) is divisible by 7 for all integer values of n greater than zero.

Proof by induction of a relation is achieved by first showing that it is true for n = 1, then if it can be assumed to be true for n it should be true for all n + 1.

If n = 1, the value of 4*6^(2n) + 3*2^(3n) is 4*6^2 + 3*2^3 = 4*36 + 3*8 = 144 + 24 = 168 and 168 is a multiple of 7.

Assume 4*6^(2n) + 3*2^(3n) is divisible by 7

4*6^(2n) + 3*2^(3n) = 7*k

4*6^(2*(n+1)) + 3*2^(3*(n+1))

= 4*6^(2n+2) + 3*2^(3n + 3)

= 4*6^(2n)*6^2 + 3*2^(3n)*8

= 4*6^(2n)*36 + 3*2^(3n)*8

= 4*6^(2n)*(35 + 1) + 3*2^(3n)*(7 + 1)

= 4*6^(2n)*(35) + 4*6^(2n) + 3*2^(3n)*(7) + 3*2^(3n)

= 4*6^(2n)*(35) + + 3*2^(3n)*(7) + 4*6^(2n) + 3*2^(3n)

The first 2 terms are divisible by 7 as both 35 and 7 are multiples of 7 and it has been assumed that the sum 4*6^(2n) + 3*2^(3n) is divisible by 7.

This proves that 4*6^(2n) + 3*2^(3n) is divisible by 7.