In the expansion of (x+a)n, sum of the odd terms is P and the sum of the even terms is Q,then 4PQ=
A
(x2−a2)n
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B
(x+a)2n+(x−a)2n
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C
(x+a)2n−(x−a)2n
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D
(x−a)2n−(x+a)2n
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Solution
The correct option is A(x+a)2n−(x−a)2n 4PQ=(P+Q)2−(P−Q)2...(i) Now P+Q= sum of all coefficients =(x+a)n...(a) P−Q implies even terms are negative, ie, alternate positive and negative terms =(x−a)n...(b) Substituting a and b in Eq (i) we get 4PQ=(x+a)2n−(x−a)2n