Given:

P (n) = “2^{n} ≥ 3n” and p(r) is true.

We have, P (n) = 2^{n} ≥ 3n

Since, P (r) is true

So,

2^{r}≥ 3r

Now, let’s multiply both sides by 2

2×2^{r}≥ 3r×2

2^{r + 1}≥ 6r

2^{r + 1}≥ 3r + 3r [since 3r>3 = 3r + 3r≥3 + 3r]

∴ 2^{r + 1}≥ 3(r + 1)

Hence, P (r + 1) is true.

Answered by Aaryan | 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]

7^{2n} + 2^{3n – 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.

(ab)^{ n} = a^{n} b^{n} for all n ϵ N