We can write the given statement as

P(n): (2n +7) < (n + 3)2

If n = 1 we get

2.1 + 7 = 9 < (1 + 3)^{2} = 16

Which is true.

Consider P (k) be true for some positive integer k

(2k + 7) < (k + 3)^{2} … (1)

Now let us prove that P (k + 1) is true.

Here

{2 (k + 1) + 7} = (2k + 7) + 2

We can write it as

= {2 (k + 1) + 7}

From equation (1) we get

(2k + 7) + 2 < (k + 3)^{2} + 2

By expanding the terms

2 (k + 1) + 7 < k^{2} + 6k + 9 + 2

On further calculation

2 (k + 1) + 7 < k^{2} + 6k + 11

Here k^{2} + 6k + 11 < k^{2} + 8k + 16

We can write it as

2 (k + 1) + 7 < (k + 4)^{2}

2 (k + 1) + 7 < {(k + 1) + 3}^{2}

P (k + 1) is true whenever P (k) is true.

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

Answered by Abhisek | 1 year agoGiven an example of a statement P (n) such that it is true for all n ϵ N.

a + (a + d) + (a + 2d) + … + (a + (n-1)d) = \( \dfrac{n}{2}\) [2a + (n-1)d]

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n (n + 1) (n + 5) is a multiple of 3 for all n ϵ N.

(ab)^{ n} = a^{n} b^{n} for all n ϵ N