Question

Find n in the binomial `(root(3)(2) + 1/(root(3)(3)))^n` if the ratio of 7^th term from the beginning to the 7^th term from the end is `1/6`

Answer 1

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

= `(2^(1/3) + 1/3^(1/3))^"n"`

General Term `"T"(r + 1) = ""^n"C"_r x^(n - r) y^r`

T⁷ = T₆₊₁ = `""^n"C"_6 (2^(1/3))^(n - 6)  (1/(3^(1/3)))^6`

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

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

7^th term from the end = (n – 7 + 2)^th term from the beginning

= (n – 5)^th term from the beginning

So, `"T"_(n - 6 + 1) = ""^n"C"_(n - 6) (2^(1/3))^(n - n + 6) (1/3^(1/3))^(n - 6)`

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

= `""^n"C"_(n - 6) (2)^2 (3)^((6 - n)/3)`

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

⇒ `(""^n"C"_(n - 6) (2)^((n - 6)/3) (3)^-2)/(""^n"C"_(n - 6) (2)^2 (3)^((6 - n)/3)) = 1/6`

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

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

⇒ `(2)^((n - 12)/3) * (3)^((n - 12)/3)` = (6)^-1

⇒ `(6)^((n - 12)/3) = (6)^-1`

⇒ `(n - 12)/3` = – 1

⇒ n – 12 = – 3

⇒ n = 12 – 3 = 9

Hence, the required value of n is 9.