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Question

Let $$P(n)$$ be the statement $$2^{n}<n!$$ where $$n$$ is a natural number, then $$P(n)$$ is true for:


A
all n
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B
all n>2
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C
all n>3
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D
all n<3
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Solution

The correct option is C all $$n>3$$

We have,

$$P\left( n \right)$$ be the statement $${{2}^{n}}<n!$$

Where $$n$$ is a natural number

Then,

Put $$n=1,2,3....$$

So,

$$P\left( 1 \right)\,$$ be the statement of $${{2}^{1}}<1!=2<1\,\,\,\left( \text{It}\,\text{is}\,\text{wrong} \right)$$

$$P\left( 2 \right)$$  be the statement of $${{2}^{2}}<2!=4<2\,\,\left( \text{It}\,\text{is}\,\text{wrong} \right)$$

$$P\left( 3 \right)$$  be the statement of $${{2}^{3}}<3!=8<6\,\,\left( \text{It}\,\text{is}\,\text{wrong} \right)$$


But,

$$P\left( 4 \right)$$  be the statement of $${{2}^{4}}<4!=16<24\,\,\left( \text{It}\,\text{is}\,\text{right} \right)$$

Similarly,

$$P\left( 5 \right),\,P\left( 6 \right)\,.......$$ So, all number $$n$$ is a natural number.


Therefore, it is true for all $$n>3$$.


Hence, this is the answer.


Maths

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