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In the Expansion (\sqrt{3}+1)^{n}, the ratio of the 7th term from the beginning to the 7th term from the end is 1: 3. Then, n is equal to

Option: 1

10


Option: 2

7


Option: 3

9


Option: 4

12


Answers (1)

best_answer

In the expansion of (\sqrt{3}+1)^{n}

7th term from beginning is ^{n} C_{6}\left(3^{\frac{1}{2}}\right)^{n-6}

And 7th term from end is equal to 7th term from beginning of (1+\sqrt{3})^{n} = ^{n} C_{6}\left(3^{\frac{1}{2}}\right)^{6}

So, 

\\\frac{^{n} C_{6}\left(3^{\frac{1}{2}}\right)^{n-6}}{^{n} C_{6}\left(3^{\frac{1}{2}}\right)^{6}}=\frac{1}{3}\;\;\;\;\;\;\;\;\;\;\\\\ \Rightarrow\left(3^{\frac{1}{2}}\right)^{n-12}=3^{-1} \\\\ \Rightarrow 3^{\frac{n-12}{2}}=3^{-1}\\ \Rightarrow n-12=-2 \\ \Rightarrow n=10

hence, option A is correct

Posted by

Pankaj

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