Suppose (x+a)n=T0+T1+T2+…+Tn Then T0−T2+T4−…2+T1−T3+T5−…2 is equal to
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a
x2+a2n
b
x2−a2n
c
(x+a)2n
d
(x−a)2n
answer is A.
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Detailed Solution
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