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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.
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Solution
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)`
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