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Q.

The middle term of x32y3+3xy3213 are

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a

13C6x923y2,13C73x2y92

b

13C7x923y213C63x2y92

c

3C6x123y2,13C7

d

None

answer is A.

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Detailed Solution

General term Tr+1=Cr   nan−rbrin the expansion of (a+b)nHere n = 13 (odd)Middle term  =13+12,13+32T7=T6+1=13C6x32y313−63xy326=13C6x923y2T8=T7+1=13C7x32y313−73xy327=13C73x2y92
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