In the expansion of (1 + a)ᵐ ⁺ ⁿ, prove that the coefficients of aᵐ and aⁿ are equal

Solution:

Assuming that a^m occurs in the (r + 1)ᵗʰ term of the expansion (1 + a)ᵐ ⁺ ⁿ, we obtain

Tᵣ ₊ ₁ = ᵐ ⁺ ⁿCᵣ (1)ᵐ ⁺ ⁿ ⁻ ʳ (a)ʳ

Comparing the exponents of a in aᵐ and in Tᵣ ₊ ₁, we obtain r = m

Therefore, the coefficient of aᵐ is

ᵐ ⁺ ⁿCₘ = [(m + n)! / (m!(m + n - m)!) ]

= (m + n)! / ((m!)(n!)) ....(1)

Assuming that aⁿ occurs in the (k + 1)ᵗʰ term of the expansion (1 + a)ᵐ ⁺ ⁿ, we obtain

Tₖ ₊ ₁ = ᵐ ⁺ ⁿCₖ (1)ᵐ ⁺ ⁿ ⁻ ᵏ (a)ᵏ

Comparing the exponents of a in aⁿ and in Tₖ ₊ ₁, we obtain k = n

Therefore, the coefficient of aⁿ is

ᵐ ⁺ ⁿCₙ = [(m + n)! / (n!(m + n - n)!) ]

= (m + n)! / ((m!)(n!))  ....(2)

Thus from (1) and (2), it is clear that the coefficients of aᵐ and aⁿ in the expansion of (1 + a)ᵐ ⁺ ⁿ are equal.

Hence proved

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NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.2 Question 9

In the expansion of (1 + a)ᵐ ⁺ ⁿ, prove that the coefficients of aᵐ and aⁿ are equal

Summary:

Using the binomial theorem, we have have proved that coefficients of aᵐ and aⁿ in the expansion of (1 + a)ᵐ ⁺ ⁿ are equal