(p + 2)th term from the end in the binomial expansion ofx2−2x22n+1 is

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2n+1C2n−n(−2)2n−px2p+1−2n

2n+1C2n−n(−2)2n−px2n−2p

2n+1C2n−p(−2)2n−px2n−2p+1

2n+1C2n−p(−2)2n−px2p−1+2n

answer is A.

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Detailed Solution

(p + 2)th term from the end=[2n+2−(p+1)]th=(2n−p+1)thfrom the beginning and T2n−p+1=2n+1C2n−px22n+1−(2n−p)−2x22n−p=2n+1C2n−p(−2)2n−px2p+1−2n

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