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  • Find n in the binomial cube root of 2 plus 1 divided cube root of 3 exponent of n if the ratio of 7th term from the beginning to the 7th term from the end is 1 divided 6

Find n in the binomial \left (\sqrt[3]{2} +\frac{1}{\sqrt[3]{3}} \right )^{n} if the ratio of 7th term from the beginning to the 7th term from the end is \frac{1}{6}

Answers (1)

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

Now, from the beginning,

Seventh term is T_{7}=T_{6+1}

                                  =^{n}C_{6} \left (\sqrt[3]{2} \right )^{n-6}\left ( \frac{1}{\sqrt[3]{3}} \right )^{6}...................(i)

& from the end,

The seventh term is the same ,i.e., \left (\frac{1}{\sqrt[3]{3}}+\sqrt[2]{3} \right )^{n}

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

Now, it is given that

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

Thus 

\frac{\left ( \sqrt[3]{2} \right )^{n-12}}{\left (\frac{1}{\sqrt[3]{3}} \right )^{n-12} }=\frac{1}{6}

Thus 

( \sqrt[3]{2} \sqrt[3]{3})^{n-12}=6^{-1}

Thus 6^\frac{n-12}{3}=6^{-1}

\frac{n-12}{3}=-1

Thus n=9

Posted by

infoexpert22

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