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Which of the following statements is true for all positive integers n ?

 

Option: 1

{{2}^{n}}>{{n}^{2}}


Option: 2

{{n}^{3}}<{{3}^{n}}


Option: 3

n{ !\,}> 2^{n}


Option: 4

{{n}^{2}}>2n


Answers (1)

best_answer

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

For n = 1, we have,

{{1}^{3}}=1<{{3}^{1}}=3

 which is true. 

Therefore,

{{n}^{3}}<{{3}^{n}} is true for n = 1.

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

\therefore {{k}^{3}}<{{3}^{k}}

We need to prove that the statement is also true for  (k+1) i.e. {{(k+1)}^{3}}<{{3}^{(k+1)}}

Expanding {{(k+1)}^{3}} we get:

{{(k+1)}^{3}}={{k}^{3}}+3{{k}^{2}}+3k+1

Now, using the inductive hypothesis

{{k}^{3}}<{{3}^{k}}. we can say:

{{k}^{3}}+3{{k}^{2}}+3k+1<{{3}^{k}}+3{{k}^{2}}+3k+1

To prove that {{(k+1)}^{3}}<{{3}^{(k+1)}}  it suffices to show that

i.e, {{3}^{k}}+3{{k}^{2}}+3k+1<{{3}^{(k+1)}} 

3{{k}^{2}}+3k<2({{3}^{k}})-1

We can see that this inequality holds for all positive integers k. 

Therefore, the inductive step is proved. 

Hence, by the principle of mathematical induction, we can say that 

{{n}^{3}}<{{3}^{n}} 

 for all positive integers n.

 

 

Posted by

manish painkra

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