If the seventh term from the beginning and end in the binomial expansion of `(2 3+1/(3 3))^n ,""` are equal, find `ndot`

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)`
and 7th term from end i.e., `T_(7)` from the beginning of `((1)/(root(3)(3)) + root(3)(2))^(n)`
i.e., `T_(7) = .^(n)C_(6) ((1)/(root(3)(3)))^(n - 6) (root (3)(2))^(6)`
given that, `(.^(n)C_(6) (root(3)(2))^(n - 6) ((1)/(root(3)(3)))^(6))/(.^(n)C_(6) ((1)/(root(3)(3)))^(n - 6) (root (3)(2))^(6)) = (1)/(6) rArr (2^((n - 6)/(3)).3^(-6//3))/(3^(-((n - 6)/(3))).2^(6//3)) = (1)/(6)`
`rArr (2^((n - 6)/(3)).2^((-6)/(3))) (3^((-6)/(3)).3^((n - 6)/(3))) = 6^(-1)`