In `(3 3+1/(3 3))^n` if the ratio of 7th term from the beginning to the 7th term from the end is 1/6, then find the value of `ndot`

Options

A. `6`
B. `9`
C. `12`
D. `15`

Answer

B

Explanation

`(b)` `T_(r+1)=^(n)C_(r )a^(n-r)*b^(r )` where `a=2^(1//3)` and `b=3^(-1//3)`
`T_(7)` from beginning `=^(9)C_(6)a^(n-6)b^(6)` and `T_(7)` from end `=^(n)C_(6)b^(n-6)a^(6)`
`implies(a^(n-12))/(b^(n-12))=(1)/(6)`
`implies2^((n-12)/(3))*3^((n-12)/(3))=6^(-1)`
`n-12=-3impliesn=9`