Q. 95.0( 8 Votes )
For every positiv
Let P(n) = 7n – 3n
For n = 1,
P (1) = 71 - 31
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.
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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