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Let P(n) be is a multiple of 6 for all n∈N.
P(1): is a multiple of 6.
So, P(1) is true.
Now,we assume that P(k) is true.
P(k): is a multiple of 6
[for some constant m] ...(i)
Now, we show that P(k+1) is true.
P(k+1): is a multiple of 6.
which is clearly a multiple of 6.
Hence, by the principle of mathematical induction, given statement P(n) is true for all n ∈N