( 2nC0)2−( 2nC1)2+( 2nC2)2+⋯+( 2nC2n)2=

( 2nC0)2−( 2nC1)2+( 2nC2)2+⋯+( 2nC2n)2(1+x)2n= 2nC0+ 2nC1x+ 2nC2x2+⋯+ 2nC2nx2n(x−1)2n= 2nC0x2n− 2nC1x2n−1+ 2nC2x2n−2−⋯+(−1)2n 2nC2nx2ncoefficient of 2nCn(−1)n=( 2nC0)2−( 2nC1)2+( 2nC2)2+⋯

( 2nC0)2−( 2nC1)2+( 2nC2)2+⋯+( 2nC2n)2=

( 2nC0)2−( 2nC1)2+( 2nC2)2+⋯+( 2nC2n)2(1+x)2n= 2nC0+ 2nC1x+ 2nC2x2+⋯+ 2nC2nx2n(x−1)2n= 2nC0x2n− 2nC1x2n−1+ 2nC2x2n−2−⋯+(−1)2n 2nC2nx2ncoefficient of 2nCn(−1)n=( 2nC0)2−( 2nC1)2+( 2nC2)2+⋯

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${(^{2 n} C_{0})}^{2} - {(^{2 n} C_{1})}^{2} + {(^{2 n} C_{2})}^{2} + \dots + {(^{2 n} C_{2 n})}^{2} =$

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${(- 1)}^{n - 1}^{2 n} C_{n}$

$(- 1)^{2 n} C_{n}$

${(- 1)}^{n}^{(2 n + 1)} C_{n}$

${(- 1)}^{n - 1}^{2 n} C_{n + 1}$

answer is B.

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Detailed Solution

$\begin{array}{l} {(^{2 n} C_{0})}^{2} - {(^{2 n} C_{1})}^{2} + {(^{2 n} C_{2})}^{2} + \dots + {(^{2 n} C_{2 n})}^{2} \\ {(1 + x)}^{2 n} =^{2 n} C_{0} +^{2 n} C_{1} x +^{2 n} C_{2} x^{2} + \dots +^{2 n} C_{2 n} x^{2 n} \\ {(x - 1)}^{2 n} =^{2 n} C_{0} x^{2 n} -^{2 n} C_{1} x^{2 n - 1} +^{2 n} C_{2} x^{2 n - 2} - \dots + {(- 1)}^{2 n}^{2 n} C_{2 n} \\ x^{2 n} {coefficient of}^{2 n} C_{n} {(- 1)}^{n} = {(^{2 n} C_{0})}^{2} - {(^{2 n} C_{1})}^{2} + {(^{2 n} C_{2})}^{2} + \dots \end{array}$