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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.