Assertion :For n∈N, (n!)3 6,(3n)n<n!(2n)n

Using A.M>G.Mn13+23+...+n3>(13⋅23⋅33...n3)1/n⇒n∑n3>(n!)3/n⇒4nn2(n+1)2>(n!)3/n⇒[(2n(n+1))2]n>(n!)3⇒nn(2n+1)2n>(n!)3⇒(n!)3<nn(2n+1)2nHence Assertion (A) & Reason (R) both are true but Reason (R) is not proper.Ans: B

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Assertion :For n N, ( n! )^{3} < n^{n} ( {n+1}{2} )^{2n

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Assertion :For $n \in N$ , $(n!)^{3}$ $< n^{n} (\frac{n + 1}{2})^{2 n}$
Reason : $n > 6, (\frac{n}{3})^{n} < n! (\frac{n}{2})^{n}$

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Both Assertion and Reason are incorrect

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Assertion :For

n \in N

(n!)^{3}

< n^{n} (\frac{n + 1}{2})^{2 n}

Reason :

n > 6, (\frac{n}{3})^{n} < n! (\frac{n}{2})^{n}

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Text solutionVerified

Using

A . M > G . M

\frac{1 ^{3} + 2 ^{3} + ... + n ^{3}}{n} > (1^{3} \cdot 2^{3} \cdot 3^{3} ... n^{3})^{1/ n}

\Rightarrow \frac{\sum n ^{3}}{n} > (n!)^{3/ n}

\Rightarrow \frac{n ^{2} ( n + 1 ) ^{2}}{4 n} > (n!)^{3/ n}

\Rightarrow [(\frac{n ( n + 1 )}{2})^{2}]^{n} > (n!)^{3} \Rightarrow n^{n} (\frac{n + 1}{2})^{2 n} > (n!)^{3}

\Rightarrow (n!)^{3} < n^{n} (\frac{n + 1}{2})^{2 n}

Hence Assertion (A) & Reason (R) both are true but Reason (R) is not proper.

Ans: B

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Question Text	Assertion :For $n \in N$ , $(n!)^{3}$ $< n^{n} (\frac{n + 1}{2})^{2 n}$ Reason : $n > 6, (\frac{n}{3})^{n} < n! (\frac{n}{2})^{n}$
Answer Type	Text solution:1
Upvotes	150

Assertion :For n∈N, (n!)3 <nn(2n+1​)2n Reason :n>6,(3n​)n<n!(2n​)n

Text solutionVerified

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Assertion :For $n \in N$ , $(n!)^{3}$ $< n^{n} (\frac{n + 1}{2})^{2 n}$
Reason : $n > 6, (\frac{n}{3})^{n} < n! (\frac{n}{2})^{n}$