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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.

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Solution

## In the binomail expansion of ${\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)}^{n}$, ${\left[\left(n+1\right)-7+1\right]}^{\mathrm{th}}$ i.e., (n − 5)th term from the beginning is the 7th term from the end. Now, ${T}_{7}{=}^{n}{C}_{6}{\left(\sqrt[3]{2}\right)}^{n-6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^{6}{=}^{n}{C}_{6}×{2}^{\frac{n}{3}-2}×\frac{1}{{3}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}{T}_{n-5}{=}^{n}{C}_{n-6}{\left(\sqrt[3]{2}\right)}^{6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^{n-6}{=}^{n}{C}_{6}×{2}^{2}×\frac{1}{{3}^{\frac{n}{3}-2}}$ It is given that, ${T}_{7}={T}_{n-5}\phantom{\rule{0ex}{0ex}}{⇒}^{n}{C}_{6}×{2}^{\frac{n}{3}-2}×\frac{1}{{3}^{2}}{=}^{n}{C}_{6}×{2}^{2}×\frac{1}{{3}^{\frac{n}{3}-2}}\phantom{\rule{0ex}{0ex}}⇒\frac{{2}^{\frac{n}{3}-2}}{{2}^{2}}=\frac{{3}^{2}}{{3}^{\frac{n}{3}-2}}\phantom{\rule{0ex}{0ex}}⇒{\left(6\right)}^{\frac{n}{3}-2}={6}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{n}{3}-2=2\phantom{\rule{0ex}{0ex}}⇒n=12$ Hence, the value of n is 12.

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