Q. 255.0( 3 Votes )

# Prove the following by the principle of mathematical induction:72n + 23n – 3 . 3n – 1 is divisible by 25 for all n ϵ N

Let P(n): 72n + 23n - 3.3n - 1 is divisible by 25

For n=1

= 72 + 20.30

= 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 72n + 23n - 3.3n - 1 is divisible by 25

= 72k + 23k - 3.3k - 1 = 25λ - - - - - - - (1)

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

So, we have to show that 72k + 1 + 23k.3k is divisible by 25

= 72k + 2 + 23k.3k = 25μ

Now,

= 72(k + 1) + 23k.3k

= 72k.71 + 23k.3k

= (25λ – 23k - 3.3k - 1)49 + 23k.3k from eq 1

=

= 24×25×49λ - 23k.3k..49 + 24.23k.3k

= 24×25×49λ - 25.23k.3k

= 25(24.49λ - 23k.3k)

= 25μ

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

Hence, P(n) is true for all n N

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