Induction proof verification ##2^{n+2} < (n+1)## for all n ##\geq 6##

Therefore, initial inequality holds for all n >= 6. We're done.In summary, the statement $2^{n+2} < (n+1)!$ is proven to be true for all n $\geq 6$ by induction. The base case of n = 6 is shown to be true, and the induction step shows that the inequality holds for all subsequent values of n. Therefore, the statement is true for all n $\geq 6$.
  • #1
ciencero
1
0
$2^{n+2} < (n+1)!$ for all n $\geq 6$

Step 1: For n = 6,

$256 < 5040$.

We assume

$2^{k+2} < (k+1)!$

Induction step:

$2 * 2^{k+2} < 2*(k+1)!$

By noting $2*(k+1)! < (k+2)!$

Then $2^{k+3} < (k+2)!$
 
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  • #2
Looks fine.
n=6 is not the first place where the inequality is true, by the way.
 
  • #3
ciencero said:
$2^{n+2} < (n+1)!$ for all n $\geq 6$

Step 1: For n = 6,

$256 < 5040$.

We assume

$2^{k+2} < (k+1)!$

Induction step:

$2 * 2^{k+2} < 2*(k+1)!$

By noting $2*(k+1)! < (k+2)!$

Then $2^{k+3} < (k+2)!$
@ciencero, at this site, use double $ characters at each end for standalone LaTeX, or double # characters at each end for inline LaTeX.
 
  • #4
It seems clear the RH side will eventually dominate. LH is being multiplied by 2 from nth to (n+1)st term while RH side is being multiplied by increasingly larger factors.
 

FAQ: Induction proof verification ##2^{n+2} < (n+1)## for all n ##\geq 6##

What is an induction proof?

An induction proof is a mathematical technique used to prove that a statement or equation holds true for all values in a specific set or sequence. It involves using the principle of mathematical induction, which states that if a statement is true for a specific value (base case) and can be proven to be true for the next value (inductive step), then it must be true for all values in the set or sequence.

How do you use induction to verify the inequality ##2^{n+2} < (n+1)## for all n ##\geq 6##?

To use induction to verify this inequality, we first prove the base case, which is when n = 6. We substitute n = 6 into the inequality and show that it holds true. Then, we assume the inequality holds true for some value k, and use this assumption to prove that it also holds true for the next value, which is k+1. This completes the inductive step. By the principle of mathematical induction, we can conclude that the inequality holds true for all values of n greater than or equal to 6.

What is the base case in the induction proof for ##2^{n+2} < (n+1)##?

The base case in this induction proof is when n = 6. This means that we substitute n = 6 into the inequality and show that it holds true. This serves as the starting point for the proof.

Can the induction proof be used to verify the inequality for values of n less than 6?

No, the induction proof only guarantees that the inequality holds true for values of n greater than or equal to 6. This is because the base case is n = 6, and the inductive step only considers the next value, which is k+1. Therefore, the proof does not cover values of n less than 6.

Why is it important to specify the range of n in the inequality being verified?

It is important to specify the range of n in the inequality being verified because the validity of the inequality may not hold true for all values of n. By specifying a range, we are limiting the values of n that need to be considered in the proof, making it easier to verify the inequality. It also ensures that the inequality holds true for a specific set or sequence of numbers, rather than just a few isolated cases.

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