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7.Binomial Theorem
medium
In ${\left( {\sqrt[3]{2} + \frac{1}{{\sqrt[3]{3}}}} \right)^n}$ if the ratio of ${7^{th}}$ term from the beginning to the ${7^{th}}$ term from the end is $\frac{1}{6}$, then $n = $
A
$7$
B
$8$
C
$9$
D
None of these
Solution
(c) $\frac{1}{6} = \frac{{^n{C_6}{{({2^{1/3}})}^{n – 6}}{{({3^{ – 1/3}})}^6}}}{{^n{C_{n – 6}}{{({2^{1/3}})}^6}{{({3^{ – 1/3}})}^{n – 6}}}}$or ${6^{ – 1}} = {6^{ – 4}}{.6^{n/3}} = {6^{n/3 – 4}}$
$\frac{n}{3} – 4 = – 1$
$ \Rightarrow $ $n = 9.$
Standard 11
Mathematics