Find `n` in the binomial `(2 3+1/(3 3))^n` , if the ration 7th term from the beginning to the 7 term from the end `1/6dot`

Explanation

Here, the Binomial expansion is `( root(3) (2) + 1/(root(3)3) )^n`


Now, `7th` term from beginning `T_7 = T_(6 + 1) = . n_(C_6) ( root(3)( 2) )^(n − 6) ( 1/root(3) (3 ))^ 6 `


i.e., `T_7` from the beginning of `(1/root(3) (3) + root (3) (2))`


i.e., `T_7 = n_(C_6) ( 1/root(3) (3))^(n − 6) ( root(3) (2))^6`


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