if the seventh term from the beginning and end in the binomial expansion of
Question:

if the seventh term from the beginning and end in the binomial expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n}$ are equal, find $n$.

Solution:

In the binomail expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n},[(n+1)-7+1]^{\text {th }}$ i.e., $(n-5)^{\text {th }}$ term from the beginning is the $7^{\text {th }}$ term from the end.

Now,

$T_{7}={ }^{n} C_{6}(\sqrt[3]{2})^{n-6}\left(\frac{1}{\sqrt[3]{3}}\right)^{6}={ }^{n} C_{6} \times 2^{\frac{n}{3}-2} \times \frac{1}{3^{2}}$

And,

$T_{n-5}={ }^{n} C_{n-6}(\sqrt[3]{2})^{6}\left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}={ }^{n} C_{6} \times 2^{2} \times \frac{1}{3^{\frac{n}{3}-2}}$

It is given that,

$T_{7}=T_{n-5}$

$\Rightarrow^{n} C_{6} \times 2^{\frac{n}{3}-2} \times \frac{1}{3^{2}}=^{n} C_{6} \times 2^{2} \times \frac{1}{3^{\frac{n}{3}-2}}$

$\Rightarrow \frac{2^{\frac{n}{3}-2}}{2^{2}}=\frac{3^{2}}{3^{\frac{n}{3}-2}}$

$\Rightarrow(6)^{\frac{n}{3}-2}=6^{2}$

$\Rightarrow \frac{n}{3}-2=2$

$\Rightarrow n=12$

Hence, the value of is 12.

 

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