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प्रश्न
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] for all n ∈ N .
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उत्तर
Let P(n) be the given statement.
Thus, we have .
\[P\left( n \right): \frac{\left( 2n \right)!}{2^{2n} \left( n! \right)^2} \leq \frac{1}{\sqrt{3n + 1}}\]
\[\text{ Step1} : \]
\[P(1): \frac{2!}{2^2 . 1} = \frac{1}{2} \leq \frac{1}{\sqrt{3 + 1}}\]
\[\text{ Thus, P(1) is true} . \]
\[\text{ Step2: } \]
\[\text{ Let P(m) be true .} \]
\[\text{ Thus, we have: } \]
\[\frac{\left( 2m \right)!}{2^{2m} \left( m! \right)^2} \leq \frac{1}{\sqrt{3m + 1}}\]
\[\text{ We need to prove that P(m + 1) is true .} \]
Now,
\[P(m + 1): \]
\[\frac{(2m + 2)!}{2^{2m + 2} \left( (m + 1)! \right)^2} = \frac{\left( 2m + 2 \right)\left( 2m + 1 \right)\left( 2m \right)!}{2^{2m} . 2^2 \left( m + 1 \right)^2 \left( m! \right)^2}\]
\[ \Rightarrow \frac{(2m + 2)!}{2^{2m + 2} \left( (m + 1)! \right)^2} \leq \frac{\left( 2m \right)!}{2^{2m} \left( m! \right)^2} \times \frac{\left( 2m + 2 \right)\left( 2m + 1 \right)}{2^2 \left( m + 1 \right)^2}\]
\[ \Rightarrow \frac{(2m + 2)!}{2^{2m + 2} \left( (m + 1)! \right)^2} \leq \frac{2m + 1}{2\left( m + 1 \right)\sqrt{3m + 1}}\]
\[\Rightarrow \frac{\left( 2m + 2 \right)!}{2^{2m + 2} \left( \left( m + 1 \right)! \right)^2} \leq \sqrt{\frac{\left( 2m + 1 \right)^2}{4 \left( m + 1 \right)^2 \left( 3m + 1 \right)}}\]
\[ \Rightarrow \frac{\left( 2m + 2 \right)!}{2^{2m + 2} \left( \left( m + 1 \right)! \right)^2} \leq \sqrt{\frac{\left( 4 m^2 + 4m + 1 \right) \times \left( 3m + 4 \right)}{4\left( 3 m^3 + 7 m^2 + 5m + 1 \right)\left( 3m + 4 \right)}}\]
\[ \Rightarrow \frac{\left( 2m + 2 \right)!}{2^{2m + 2} \left( \left( m + 1 \right)! \right)^2} \leq \sqrt{\frac{12 m^3 + 28 m^2 + 19m + 4}{\left( 12 m^3 + 28 m^2 + 20m + 4 \right)\left( 3m + 4 \right)}}\]
\[ \because \frac{12 m^3 + 28 m^2 + 19m + 4}{\left( 12 m^3 + 28 m^2 + 20m + 4 \right)} < 1\]
\[ \therefore \frac{\left( 2m + 2 \right)!}{2^{2m + 2} \left( \left( m + 1 \right)! \right)^2} < \frac{1}{\sqrt{3m + 4}}\]
Thus, P(m + 1) is true.
Hence, by mathematical induction
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] is true for all n ∈ N
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