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Suppose $(x + a)^{n} = T_{0} + T_{1} + T_{2} + \dots + T_{n}$ Then ${(T_{0} - T_{2} + T_{4} - \dots)}^{2} + {(T_{1} - T_{3} + T_{5} - \dots)}^{2}$ is equal to

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x2+a2n

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(x−a)2n

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detailed solution

Correct option is A

Not that Tr=nCrxn−rar.(x+ai)n=nC0xn+nCrxn−1(ia)+nC2xn−2(ia)2+nC3xn−3(ia)3+…+nCn(ia)n=T0−T2+T4−…+iT¯1−T¯3+T¯5−…⇒x2+a2n=1(x+ai)n2=T0−T2+T4−…2+T1−T3+T5−…2

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Suppose (x+a)n=T0+T1+T2+…+Tn Then T0−T2+T4−…2+T1−T3+T5−…2 is equal to

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Suppose $(x + a)^{n} = T_{0} + T_{1} + T_{2} + \dots + T_{n}$ Then ${(T_{0} - T_{2} + T_{4} - \dots)}^{2} + {(T_{1} - T_{3} + T_{5} - \dots)}^{2}$ is equal to