If P (n) is the statement “2n ≥ 3n”, and if P (r) is true, prov – InterviewSolution

InterviewSolution

1.

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

Answer»

Given as

P (n) = “2ⁿ ≥ 3n” and p(r) is true.

We have, P (n) = 2ⁿ ≥ 3n

Here, P (r) is true

Therefore,

2^r≥ 3r

Then, 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)

Thus, P (r + 1) is true.

Discussion

No Comment Found

Related InterviewSolutions

State the first principle of mathematical induction.
The distributive law from algebra states that for real numbers c, a1 and a2, we have c(a1 + a2) = ca1 + ca2 Use this law and mathematical induction to prove that, for all natural numbers, n ≥ 2, if c, a1, a2, …... an are any real numbers, then c(a1 + a2 +…+ an) = ca1 + ca2 +…+ can.
Prove the following by the principle of mathematical induction: n(n + 1) (n + 5) is a multiple of 3 for allnϵ N. Show that: P(n): n(n + 1) (n + 5) is multiple by 3 for all n∈N
If P(n) : 49n + 16n + λ is divisible by 64 for n ∈ N is true, then the least negative integral value of λ is A. -3 B. -2 C. -1 D. -4
Prove by the principle of mathematical induction: 12 + 22 + 32 + … + n2 = [n(n + 1) (2n + 1)]/6
Prove by the principle of mathematical induction:(ab)n = an bn for all n ϵ N
Given an example of a statement P (n) such that it is true for all n ϵ N.
Prove by the principle of mathematical induction:1 + 2 + 3 + … + n = n(n +1)/2 i.e., the sum of the first n natural numbers is n(n + 1)/2.
Prove by the principle of mathematical induction: 1 + 3 + 32 + … + 3n-1 = (3n – 1)/2
If P (n) is the statement “n2 – n + 41 is prime”, prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.