If P (n) is the statement “2n ≥ 3n”, and if P (r) is true, prove that P (r + 1) is true.

Question

If P (n) is the statement “2ⁿ ≥ 3n”, and if P (r) is true, prove that P (r + 1) is true.

Asked by Sakshi | 1 year ago | 108

0 0

Class 11 Maths Principle of Mathematical Induction Add Answer

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

7²ⁿ + 2^{3n – 3}. 3n – 1 is divisible by 25 for all n ϵ N

(ab)ⁿ = aⁿ bⁿ for all n ϵ N

Accepted Answer

Given:

P (n) = “2ⁿ ≥ 3n” and p(r) is true.

We have, P (n) = 2ⁿ ≥ 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 ago