If the seventh terms from the beginning and the end in the expansion of `(root(3)(2) + 1/(root(3)(3)))^n` are equal, then n equals ______.

#### Solution

If the seventh terms from the beginning and the end in the expansion of `(root(3)(2) + 1/(root(3)(3)))^n` are equal, then n equals **12**.

**Explanation:**

The given expansion is `(root(3)(2) + 1/(root(3)(3)))^n`

∴ T_{7} = T_{6+1}

= `""^n"C"_6 (2^(1/3))^(n - 6) * 1/((3^(1/3))^6`

= `""^n"C"_6 (2)^((n - 6)/3) * 1/(3)^2`

Now the T_{7 }from the end = T_{7 } from the beginning in `(1/(root(3)(2)) + root(3)(2))^n`.

∴ T_{7} = T_{6+1 }

= `""^n"C"_6 (1/(3^(1/3)))^(n - 6) * (2^(1/3))^6`

We get `""^n"C"_6 (2)^((n - 6)/3) * (1/3^2) = ""^n"C"_6 1/((n - 6)/3) * (2)^2`

⇒ `(2)^((n - 6)/3) * (3)^-2 = (3)^-((n - 5)/3) * (2)^2`

⇒ `(2)^((n - 6)/3 2) * (3)^(-2 + (n - 6)/3)` = 1

⇒ `2^((n - 12)/3) * (3)^((n - 12)/3)` = 1

⇒ `(6)^((n - 12)/3) = (6)^0`

⇒ `(n - 12)/3` = 0

⇒ n = 12