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Question

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
(x2a2)n
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B
(x+a)2n+(xa)2n
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C
(x+a)2n(xa)2n
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D
(xa)2n(x+a)2n
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Solution

The correct option is A (x+a)2n(xa)2n
4PQ=(P+Q)2(PQ)2 ...(i)
Now P+Q= sum of all coefficients
=(x+a)n ...(a)
PQ implies even terms are negative, ie, alternate positive and negative terms
=(xa)n ...(b)
Substituting a and b in Eq (i) we get
4PQ=(x+a)2n(xa)2n

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