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 < k2 + 6k + 9 + 2
On further calculation
2 (k + 1) + 7 < k2 + 6k + 11
Here k2 + 6k + 11 < k2 + 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]
72n + 23n – 3. 3n – 1 is divisible by 25 for all n ϵ N
n (n + 1) (n + 5) is a multiple of 3 for all n ϵ N.