Q. 95.0( 8 Votes )

For every positiv

Answer :

Let P(n) = 7n – 3n


For n = 1,


P (1) = 71 - 31


= 4


Which is divisible by 4.


Thus P (1) is true.


Consider P(K) to be true for some natural number k.


i.e. P(K) = 7k – 3k is divisible by 4.


We now have to prove P(K+1) is divi9sible by 4 whenever P(K) is true.


So,


7K+1 – 3k+1 = 7K+1 – 7.3k + 7.3k+ 3k+1


= 7(7k – 3k) + (7-3)3k


= 7(4d) + (7-3)3k


= 7(4d) + 4.3k


= 4(7d + 3k)


Hence, we can see that 7K+1 – 3k+1 is divisible by 4.


Thus P(K+1) is true when P(K) is true.


By mathematical induction the given statement is true for all n.


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