Find the middle terms in the expansion of (3 - x³/6)⁷

Solution:

It is known that in the binomial expansion of (a + b)ⁿ, when n is odd; there are two middle terms, namely [(n + 1)/2]ᵗʰ term and [(n + 1)/2 + 1]ᵗʰ term.

Therefore the middle term in the expansion (3 - x³/6)⁷ are [(7 + 1)/2] = 4ᵗʰ term and [(7 + 1)/2 + 1] = 5ᵗʰ term.

T₄ = T₃ ₊ ₁

= ⁷C₃ (3)⁷ ⁻ ³ (- x³/6)³

= [7!/(3!)(4!)] x (3⁴) x (- 1)³ x (x⁹/6³)

= [- 7 x 6 x 5 x (4!)]/[3 x 2 x (4!)] x (3⁴) x (1/(2³ x 3³)) x (x⁹)

= (-105/8)x⁹

Now,

T₅ = T₄ ₊ ₁

= ⁷C₃ (3)⁷ ⁻ ⁴ (-x³/6)⁴

= [7!/(4!)(3!)] x (3³) x (x¹²/6⁴)

= (7.6.5.4!)/(3.2!(4)! x (3³) x [1/(2⁴ x 3⁴)] (x¹²)

= (35/48)x¹²

Thus, the middle terms in the expansion of (3 - x³/6)⁷ are (105/8)x⁹ and (35/48)x¹²

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NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.2 Question 7

Find the middle terms in the expansion of (3 - x³/6)⁷

Summary:

Using binomial theorem, middle terms in the expansion of (3 - x³/6)⁷ are to be written. We have found that the middle terms in the expansion of (3 - x³/6)⁷ to be (-105/8)x⁹ and (35/48)x¹²