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If ${(1 + x + x^{2})}^{48} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{96} x^{96}$ then the value of $a_{0} - a_{2} + a_{4} - a_{6} + \dots + a_{96}$

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Correct option is C

Putting x = i, we get1+i+i248=a0+a1i+a2i2+…+a98i96⇒ i48=a0−a2+a4−…+ia1−a3+…Equating the real parts we geta0−a2+a4−a6+…+a96=1

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If 1+x+x248=a0+a1x+a2x2+…+a96x96 then the value of a0−a2+a4−a6+…+a96

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If ${(1 + x + x^{2})}^{48} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{96} x^{96}$ then the value of $a_{0} - a_{2} + a_{4} - a_{6} + \dots + a_{96}$