Question

Find the sixth term of the expansion `(y^(1/2) + x^(1/3))^"n"`, if the binomial coefficient of the third term from the end is 45.

Answer 1

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

Since the binomial coefficient of third term from the end = Binomial coefficient of third term from the beginning = ⁿC₂

∴ ⁿC₂ = 45

⇒ `("n"("n" - 1))/2` = 45

⇒ n² – n = 90

⇒ n² – n – 90 = 0

⇒ n² – 10n + 9n – 90 = 0

⇒ n(n – 10) + 9(n – 10) = 0

⇒ (n – 10)(n + 9) = 0

⇒ n = 10, n = –9

⇒ n = 10, n ≠ – 9

So, the given expression becomes `(y^(1/2) + x^(1/3))^10`

Sixth term is this expression T₆ = T₅₊₁

= `""^10"C"_5 (y^(1/2))^(10 - 5)  (x^(1/3))^5`

= `""^10"C"_5  y^(5/2) * x^(5/3)`

= `252  y^(5/2) x^(5/3)`

Hence, the required term = `252  y^(5/2) * x^(5/3)`