Find the middle term in the expansion of (x/3 + 9y)10

1.	Find the middle term in the expansion of (x/3 + 9y)10
Answer» x = x/3, a = 9y, n = 10, r = 5 n = 10 (even) there is only one middle term is (n/2 + 1) = (10/2 + 1) = 6^th term General term T_r+1 = ⁿC_r x^n-r a^r⇒ T_s+1 = ¹⁰C₅ (x/3)^{10 - 5}(9y)⁵ T₆= ¹⁰C₅ (x/3)⁵ g5 y5 ⇒ T6 = 10C5 . x⁵/3⁵ × 9⁵ y⁵ = ¹⁰C₅ (xy)5 × (3¹⁰/3⁵) T₆ = [¹⁰C₅ (xy)⁵ 3⁵] is the middle term.

Answer»

x = x/3, a = 9y, n = 10, r = 5

n = 10 (even) there is only one middle term is (n/2 + 1) = (10/2 + 1) = 6^th term

General term T_r+1 = ⁿC_r x^n-r a^r⇒ T_s+1 = ¹⁰C₅ (x/3)^{10 - 5}(9y)⁵

T₆= ¹⁰C₅ (x/3)⁵ g5 y5 ⇒ T6 = 10C5 . x⁵/3⁵ × 9⁵ y⁵ = ¹⁰C₅ (xy)5 × (3¹⁰/3⁵)

T₆ = [¹⁰C₅ (xy)⁵ 3⁵] is the middle term.

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Find the middle term in the expansion of (x/3 + 9y)10

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