Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)²ⁿ ⁻ ¹

Solution:

It is known that (r + 1)ᵗʰ term, (Tᵣ ₊ ₁) in the binomial expansion of (a + b)ⁿ is given by Tᵣ ₊ ₁ = ⁿCᵣ aⁿ ⁻ ʳ bʳ

Assuming that xⁿ occurs in the (r + 1)ᵗʰ term of the expansion (1 + x)²ⁿ, we obtain

Tᵣ ₊ ₁ = ²ⁿCᵣ (1)² ⁻ ʳ (x)ʳ

= ²ⁿCᵣ xʳ

Comparing the indices of x in x and in Tᵣ ₊ ₁, we obtain r = n.

Therefore, the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is ²ⁿCₙ.

Using nCr formula,

²ⁿCₙ = [(2n)!/n!(2n - n)!]

= [(2n)!/(n!)(n!)]

= [(2n)!/(n!)]² ....(1)

Assuming that xⁿ occurs in the (k + 1)ᵗʰ term of the expansion (1 + x)²ⁿ⁻ ¹, we obtain

Tₖ ₊ ₁ = ²ⁿ⁻ ¹Cₖ (1)²ⁿ ⁻ ¹ ⁻ ᵏ (x)ᵏ

= ²ⁿ⁻ ¹Cₖ xᵏ

Comparing the indices of x in x and in Tₖ ₊ ₁, we obtain k = n.

Therefore, the coefficient of xⁿ in the expansion of (1 + x)²ⁿ⁻ ¹ is ²ⁿ⁻ ¹Cₙ.

Using nCr formula,

²ⁿ ⁻ ¹Cₙ = [(2n - 1)!/n!(2n - 1 - n)!]

= [(2n - 1)!/(n!)(2n - 1 - n!)]

= [(2n - 1)!/n!(n - 1)!]

Multiplying and dividing by 2n,

= [2n × (2n - 1)! / 2n × n!(n - 1)!]

= (2n)! / [2 × (n!)(n!)]

= 1/2 [(2n)!/(n!)²] ....(2)

From (1) and (2), we can observe that

²ⁿ ⁻ ¹Cₙ = 1/2 (²ⁿCₙ)

²ⁿCₙ = 2 (²ⁿ ⁻ ¹Cₙ)

Therefore, the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)²ⁿ ⁻ ¹.

Hence Proved

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NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.2 Question 11

Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)²ⁿ ⁻ ¹

Summary:

Using the binomial theorem, we proved that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of 1 - x²ⁿ ⁻ ¹