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Let P(n) be the statement that 2^{n}> n{\displaystyle !}

Which of the following statements is true using mathematical induction? 

 

Option: 1

P(4) is true and for all n\geq 4, if P(n) is true, then P(n+1) is true.

 


Option: 2

P(4) and P(5) are true, and for all n\geq 5, if P(n) is true, then P(n+1)is true.


Option: 3

P(4) and P(6) are true, and for all n\geq 6, if P(n) is true, then P(n+1) is true.


Option: 4

P(4) and P(7) are true, and for all n\geq 7, if P(n) is true, then P(n+1) is true.


Answers (1)

best_answer

We will use the principle of mathematical induction to solve this problem.

When n = 4, we have,

2^{4}=\ 16 and 4{\displaystyle !\,}=24

So, 2^{4}< 4{\displaystyle !\,}

Therefore,

2^{n}< n{\displaystyle !\,}  is true for n = 4.

Assume that the statement is true for some arbitrary positive integer k.

\therefore 2^{n}< n{\displaystyle !\,}

We need to prove that the statement is also true for (k+1).

i,e,.2^{k+\ 1}< (k+\ 1){\displaystyle !\,}

Take the left-hand side of above equation.

2^{k+1}=2\left( 2k \right)

\Rightarrow 2^{k+1}< 2k{\displaystyle !\,}

\Rightarrow 2^{k+1}< \left( k+1 \right)k{\displaystyle !\,}

Thus, we have shown that P(n) is true for n=k+1.

Therefore, by the Principle of Mathematical Induction, we have proven that P(n) is true for all integers n\geq 4 So, option (A) is true, which says that P(4) is true and for all n\geq 4, if P(n) is true, then P(n+1) is true. 

Options (B), (C), and (D) have additional statements that are not necessarily true, such as P(5), P(6), or P(7) being true, which are not required to prove the original statement

Posted by

Divya Prakash Singh

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