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प्रश्न
Prove that: 4nC2n : 2nCn = [1 · 3 · 5 ... (4n − 1)] : [1 · 3 · 5 ... (2n − 1)]2.
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उत्तर
\[\frac{{}^{4n} C_{2n}}{{}^{2n} C_n} = \frac{1 . 3 . 5 . . . \left( 4n - 1 \right)}{\left[ 1 . 3 . 5 . . . \left( 2n - 1 \right) \right]^2}\]
\[LHS = \frac{{}^{4n} C_{2n}}{{}^{2n} C_n}\]
\[ = \frac{\left( 4n \right)!}{\left( 2n \right)!\left( 2n \right)!} \times \frac{n!n!}{\left( 2n \right)!}\]
\[ = \frac{\left[ 4n \times \left( 4n - 1 \right) \times \left( 4n - 2 \right) \times \left( 4n - 3 \right) . . . . . . . . . . . . . . . . . 3 \times 2 \times 1 \right] \times \left( n! \right)^2}{\left[ 2n \times \left( 2n - 1 \right) \times \left( 2n - 2 \right) . . . . . . . 3 \times 2 \times \times 1 \right]^2 \left( 2n \right)!}\]
\[ = \frac{\left[ 1 \times 3 \times 5 . . . . . . . . \left( 4n - 1 \right) \right]\left[ 2 \times 4 \times 6 . . . . . . . . . . . . . . . 4n \right] \times \left( n! \right)^2}{\left[ 1 \times 3 \times 5 \times . . . . . . . . . \left( 2n - 1 \right) \right]^2 \left[ 2 \times 4 \times 6 \times . . . . . . . 2n \right]^2 \times \left( 2n \right)!}\]
\[ = \frac{\left[ 1 \times 3 \times 5 . . . . . . . . \left( 4n - 1 \right) \right] \times 2^{2n} \times \left[ 1 \times 2 \times 3 . . . . . . . . . . 2n \right] \left( n! \right)^2}{\left[ 1 \times 3 \times 5 \times . . . . . . . . . \left( 2n - 1 \right) \right]^2 \times 2^{2n} \left[ 1 \times 2 \times 3 \times . . . . . . . n \right]^2 \left( 2n \right)!}\]
\[ = \frac{\left[ 1 \times 3 \times 5 . . . . . . . . \left( 4n - 1 \right) \right]\left( 2n \right)! \left[ n! \right]^2}{\left[ 1 \times 3 \times 5 \times . . . . . . . . . \left( 2n - 1 \right) \right]^2 \left[ n! \right]^2 \left( 2n \right)!}\]
\[ = \frac{\left[ 1 \times 3 \times 5 . . . . . . . . \left( 4n - 1 \right) \right]}{\left[ 1 \times 3 \times 5 \times . . . . . . . . . \left( 2n - 1 \right) \right]^2} = RHS\]
\[\text{Hence, proved} .\]
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