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