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Question

If the seventh term from the beginning and end in the binomial expansion of (32+133)n are equal, find n.

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Solution

Given expansions is (32+133)n

T7=T6+1=nC6(32)n6(133)6 ...(i)

Since, T7 from end is same as the T7 from beginning of (133+32)2

Then,

T7=nC6(133)n6(32)6 ...(ii)

Given that, nC6(2)n63(3)62=nC6(3)(n6)3(2)63

(2)n123=(1313)n12

which is true, when n123=0

n12=0n=12


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