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

Open in App

Solution

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

Now,
$T_{7} =^{n} C_{6} {(\sqrt[3]{2})}^{n - 6} {(\frac{1}{\sqrt[3]{3}})}^{6} =^{n} C_{6} \times 2^{\frac{n}{3} - 2} \times \frac{1}{3^{2}} And, T_{n - 5} =^{n} C_{n - 6} {(\sqrt[3]{2})}^{6} {(\frac{1}{\sqrt[3]{3}})}^{n - 6} =^{n} C_{6} \times 2^{2} \times \frac{1}{3^{\frac{n}{3} - 2}}$

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 n is 12.

Suggest Corrections

Similar questions

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

{(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})}^{n}

\frac{1}{6}

{(2^{1 / 3} + 3^{- 1 / 3})}^{n}

1 / 6

{(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})}^{n}

n = ?

{[\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}}]}^{n}

1 : 6

n