Question

If the seventh term from the beginning and the end in the expansion of ${(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}})}^n$ are equal, then the value of $\sum _{r=0}^5{}^n{C}_r$ is equal to

• A. $\frac{1}{2}(2^{12}+{}^{12}{C}_6)$
• B. $\frac{1}{2}(2^{12}-{}^{12}{C}_6)$
• C. $\frac{1}{2}(2^{11}+{}^{11}{C}_6)$
• D. $\frac{1}{2}(2^{11}-{}^{11}{C}_6)$

Answer 1

The correct option is B $\frac{1}{2}(2^{12}-{}^{12}{C}_6)$

Given, $T_7=T_{n-5}$
$\Rightarrow {}^n{C}_6{\left(\sqrt[3]{32}\right)}^{n-6}{\left(\frac{1}{\sqrt[3]{33}}\right)}^6={}^n{C}_{n-6}(\sqrt[3]{32}{)}^6{\left(\frac{1}{\sqrt[3]{33}}\right)}^{n-6}$
$\Rightarrow (\sqrt[3]{32}{)}^{n-12}=(\sqrt[3]{33}{)}^{12-n}$
This is only possible when $n=12$

Now, $\sum _{r=0}^5{}^{12}{C}_r={}^{12}{C}_0+{}^{12}{C}_1+\cdots +{}^{12}{C}_5$
$=\frac{1}{2}(2^{12}-{}^{12}{C}_6)[\because {}^{12}{C}_0+{}^{12}{C}_1+\cdots +{}^{12}{C}_{12}=2^{12}]$