In the binomial ${(2^{1 / 3} + 3^{- 1 / 3})}^{n}$ if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is $1 / 6$ then n is equal to

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Solution

The correct option is B 9
From the above question we can conclude
$6 T_{7} = T_{n - 7}$
Hence
$^{n} C_{6} 2^{\frac{n - 6}{3}} {.3}^{- 2} .6 =^{n} C_{6} 3^{\frac{6 - n}{3}} {.2}^{2}$
Hence
$(2.3)^{\frac{n - 6}{3}} .6 = (2.3)^{2}$
Hence
$6^{\frac{n - 6}{3}} = \frac{6^{2}}{6} = 6^{1}$
Comparing the indices, we get
$n - 6 = 3$
$n = 9$ .

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In the binomial (21/3+3−1/3)n if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is 1/6 then n is equal to

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