Question

If the seventh terms from the beginning and the end in the expansion of ${(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}})}^n$ are equal then, $n=?$

• A. $9$
• B. $14$
• C. $22$
• D. $12$

Answer 1

Answer: (D)

$12$

Solution:

Seventh term from beginning i.e. $T_7$ in the expansion of ${(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}})}^n={}^n{C}_6\times 2^{\frac{6}{3}}\times 3^{\frac{-(n-6)}{3}}$

and Seventh term from end i.e. $T_{n-7+2}=T_{n-5}$ in the expansion of ${(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}})}^n={}^n{C}_{n-6}\times 2^{\frac{n-6}{3}}\times 3^{\frac{-6}{3}}$

A/q

$T_7=T_{n-5}$

or, ${}^n{C}_6\times 2^{\frac{6}{3}}\times 3^{\frac{-(n-6)}{3}}$ $={}^n{C}_{n-6}\times 2^{\frac{n-6}{3}}\times 3^{\frac{-6}{3}}$

or, $2^{\frac{-n+12}{3}}\times 3^{\frac{-n+12}{3}}=1$

or, $6^{\frac{-n+12}{3}}=6^0$

From above,

$\frac{-n+12}{3}=0$

or, $n=12$

Hence, D is the correct option.