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Question

In the binomial expansion of (32+133)n, the ratio of the 7th terms from the beginning to the 7th terms from the end is 1:6, find n.

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Solution

(32+133)n

T7=nC6(21/3)n6(31/3)6

Tfrom end=nCn8(21/3)6(31/3)n6

16=nC62n63×363nCn6 22 3n63

16=2n6363×3n6363

16=(6)n6363

61=(6)n6363

1=n6363

1+2=n63

1×3=n6
9=n

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