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Prove That: ( 2 N + 1 ) ! N ! = 2n [1 · 3 · 5 ... (2n − 1) (2n + 1)]

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प्रश्न

Prove that:

\[\frac{(2n + 1)!}{n!}\] = 2n [1 · 3 · 5 ... (2n − 1) (2n + 1)]
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उत्तर

\[\text{LHS} = \frac{\left( 2n + 1 \right)!}{n!} \]
\[ = \frac{\left( 2n + 1 \right)\left( 2n \right)\left( 2n - 1 \right) . . . . \left( 4 \right)\left( 3 \right)\left( 2 \right)\left( 1 \right)}{n!}\]
\[ = \frac{\left[ \left( 1 \right)\left( 3 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right)\left( 2n + 1 \right) \right]\left[ \left( 2 \right)\left( 4 \right)\left( 6 \right) . . . . . . . . . \left( 2n \right) \right]}{n!} \]
\[ = \frac{2^n \left[ \left( 1 \right)\left( 3 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right)\left( 2n + 1 \right) \right]\left[ \left( 1 \right)\left( 2 \right)\left( 3 \right) . . . . . . . . . \left( n \right) \right]}{n!}\]
\[ = \frac{2^n \left[ \left( 1 \right)\left( 3 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right)\left( 2n + 1 \right) \right]\left[ n! \right]}{n!}\]
\[ = 2^n \left[ \left( 1 \right)\left( 3 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right)\left( 2n + 1 \right) \right] = \text{RHS}\]
\[ \text{Hence, proved} . \]

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Factorial N (N!) Permutations and Combinations
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अध्याय 16: Permutations - Exercise 16.1 [पृष्ठ ५]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 16 Permutations
Exercise 16.1 | Q 12 | पृष्ठ ५

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