First slide

Binomial theorem for positive integral Index

Question

If the ratio of the 7th term from the beginning to the 7th term from the end in the expansion of $(2^{1 / 3} +$ ${3^{- 1 / 3})}^{n}$ is $1 / 6$ then the value of $n$ is

Moderate

A

9
B

12
C

6
D

3

Solution

Let $a = 7^{th}$ term from the beginning
$= T_{7} = T_{6 + 1} =^{n} C_{6} {(2^{1 / 3})}^{n - 6} {(3^{- 1 / 3})}^{6}$
and let $b = 7^{th}$ term from the end
$= (n + 2 - 7) th or (n - 5) th$ term from the beginning
$\begin{array}{l} =^{n} C_{n - 6} {(2^{1 / 3})}^{n - (n - 6)} {(3^{- 1 / 3})}^{n - 6} \\ =^{n} C_{6} {(2^{1 / 3})}^{6} {(3^{- 1 / 3})}^{n - 6} \end{array}$
we are given
$\begin{array}{l} \frac{a}{b} = \frac{1}{6} \\ \frac{{(2^{1 / 3})}^{n - 6} {(3^{- 1 / 3})}^{6}}{{(2^{1 / 3})}^{6} {(3^{- 1 / 3})}^{n - 6}} = \frac{1}{6} \\ \Rightarrow {(\frac{(2^{1 / 3})}{(3^{- 1 / 3})})}^{n - 12} = \frac{1}{6} \\ \Rightarrow 6^{(n - 12) / 3} = 6^{- 1} \\ \Rightarrow n - 12 = - 3 \Rightarrow n = 9 . \end{array}$

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