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Sum

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 = ^{n}C_{2}

∴ ^{n}C_{2} = 45

⇒ `("n"("n" - 1))/2` = 45

⇒ n^{2} – n = 90

⇒ n^{2} – n – 90 = 0

⇒ n^{2} – 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 T_{6} = T_{5+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)`

Concept: Binomial Theorem for Positive Integral Indices

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