Prove that 2n < (n + 2)! for all n belongs to natural numbers by Principle Of Mathematical Induction.

Open in App

Solution

$Let p (n) : 2 n < (n + 2)! For n = 1 2 (1) < (1 + 2)! 2 < 3! 2 < 6 p (n) is true for n = 1 Let us assume that p (n) is true for n = k . So, 2 k < (k + 2)! Now, for n = k + 1 2 (k + 1) = 2 k + 2 < (k + 2)! + 2 < (k + 3)! (As (k + 3)! = (k + 3) (k + 2)! which is > (k + 2)! + 2) Hence by principle of mathematical induction p (n) is true for all value of n \in N .$

Suggest Corrections

Similar questions

Using principle of mathematical induction, prove that \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{n}} for all natural numbers n \geq 2 .

[NCERT EXEMPLAR]

Q. Prove by the principle of mathematical induction that

2^{n} > n

for all

n \in N .

Prove by the principle of mathematical induction that $1 \times 1! + 2 \times 2! + 3 \times 3! + . . . + n \times n! = (n + 1)! - 1$ for all natural numbers n.

Prove by using the principle of mathematical induction $\forall n \in N$

$2 + 5 + 8 + 11 + . . . + (3 n - 1) = \frac{1}{2} n (3 n + 1)$

Using principle of mathematical induction, prove that $4^{n} + 15 n - 1$ is divisible by 9 for all natural numbers n.

Prove by using the principle of mathematical induction $\forall n \in N$

$2 + 5 + 8 + 11 + . . . + (3 n - 1) = \frac{1}{2} n (3 n + 1)$

Using principle of mathematical induction, prove that $4^{n} + 15 n - 1$ is divisible by 9 for all natural numbers n.

Join BYJU'S Learning Program

Prove that 2n < (n + 2)! for all n belongs to natural numbers by Principle Of Mathematical Induction.

Let pn: 2n<n+2!For n=121<1+2!2<3!2<6pn is true for n=1 Let us assume that pn is true for n=k.So,2k<k+2!Now, for n=k+12k+1=2k+2<k+2!+2<k+3! As k+3!=k+3k+2! which is >k+2!+2Hence by principle of mathematical inductionpn is true for all value of n∈N.