Q. 255.0( 3 Votes )

# Prove the following by the principle of mathematical induction:

7^{2n} + 2^{3n – 3} . 3n – 1 is divisible by 25 for all n ϵ N

Answer :

Let P(n): 7^{2n} + 2^{3n - 3}.3^{n - 1} is divisible by 25

For n=1

= 7^{2} + 2^{0}.3^{0}

= 49 + 1

= 50

Therefor it is divisible by 25

So, P(n) is true for n = 1

Now, P(n) is true For n = k,

So, we have to show that 7^{2n} + 2^{3n - 3}.3^{n - 1} is divisible by 25

= 7^{2k} + 2^{3k - 3}.3^{k - 1} = 25λ - - - - - - - (1)

Now, P(n) is true For n = k + 1,

So, we have to show that 7^{2k + 1} + 2^{3k}.3^{k} is divisible by 25

= 7^{2k + 2} + 2^{3k}.3^{k} = 25μ

Now,

= 7^{2(k + 1)} + 2^{3k}.3^{k}

= 7^{2k}.7^{1} + 2^{3k}.3^{k}

= (25λ – 2^{3k - 3}.3^{k - 1})49 + 2^{3k}.3k from eq 1

=

= 24×25×49λ - 2^{3k}.3^{k.}.49 + 24.2^{3k}.3^{k}

= 24×25×49λ - 25.2^{3k}.3^{k}

= 25(24.49λ - 2^{3k}.3^{k})

= 25μ

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true for all n ∈ N

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