# The coefficients of the (r - 1)ᵗʰ, rᵗʰ and (r + 1)ᵗʰ term in the expansion of (x + 1)ⁿ are in the ratio of 1: 3: 5 . Find n and r

**Solution:**

It is known that (k + 1)ᵗʰ term, Tₖ ₊ ₁ term of the binomial expansion of (a + b)ⁿ, is given by Tₖ ₊ ₁ = ⁿCₖ (a)ⁿ ⁻ ᵏ (b)ᵏ.

Hence, (r - 1)ᵗʰ term in the expansion of (x + 1)ⁿ is

Tᵣ ₋ ₁ = ⁿCᵣ ₋ ₂ (x)ⁿ ⁻ ⁽ʳ ⁻ ²⁾ (1)ʳ ⁻ ²

= ⁿCᵣ ₋ ₂ xⁿ ⁻ ʳ ⁺ ²

rᵗʰ term in the expansion of (x + 1)ⁿ is

Tᵣ = ⁿCᵣ ₋ ₁ (x)ⁿ ⁻ ⁽ʳ ⁻ ¹⁾ (1)ʳ ⁻ ¹

= ⁿCᵣ ₋ ₁ xⁿ ⁻ ʳ ⁺ ¹

(r + 1)ᵗʰ term in the expansion of (x + 1)ⁿ is

Tᵣ ₊ ₁ = ⁿCᵣ(x)ⁿ ⁻ ʳ (1)ʳ

= ⁿCᵣ xⁿ ⁻ ʳ

Therefore, coefficients of the (r - 1)ᵗʰ, rᵗʰ and (r + 1)ᵗʰ term in the expansion of (x + 1)ⁿ are ⁿCᵣ ₋ ₂, ⁿCᵣ ₋ ₁, and ⁿCᵣ respectively.

Since these coefficients are in the ratio of 1: 3: 5 , we obtain

(ⁿCᵣ ₋ ₂/ⁿCᵣ ₋ ₁) = 1/3 and (ⁿCᵣ ₋ ₂/ⁿCᵣ) = 3/5

Now,

(ⁿCᵣ ₋ ₂/ⁿCᵣ ₋ ₁) = [n!/(r - 2)!(n - r + 2)!] x [(r - 1)!(n - r + 1)!/n!]

1/3 = [(r - 1).(r - 2)!(n - r + 1)!]/[(r - 2)!(n - r + 2)!(n - r + 1)!]

1/3 = (r - 1)/(n - r + 2)

n - r + 2 = 3r - 3

n - 4r + 5 = 0

n = 4r - 5 ....(1)

(ⁿCᵣ ₋ ₂/ⁿCᵣ) = [n!/(r - 1)!(n - r + 1)!] x [r !(n - r)!/n!]

3/5 = [r.(r - 1)!(n - r)!]/[(r - 1)!(n - r + 1)!(n - r)!]

3/5 = r/(n - r + 1)

3n - 3r + 3 = 5r

3n - 8r + 3 = 0 ....(2)

From (1) and (2), we obtain

3(4r - 5) - 8r + 3 = 0

12r - 15 - 8r + 3 = 0

4r - 12 = 0

r = 3

Putting the value of r in (1), we obtain n

n = 4 x 3 - 5

= 12 - 5

= 7

Thus, n = 7 and r = 3

NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.2 Question 10

## The coefficients of the (r - 1)ᵗʰ, rᵗʰ and (r + 1)ᵗʰ term in the expansion of (x + 1)ⁿ are in the ratio of 1: 3: 5 . Find n and r.

**Summary:**

Using the binomial theorem, if coefficients of the (r - 1)ᵗʰ, rᵗʰ and (r + 1)ᵗʰ term in the expansion of (x + 1)ⁿ are in the ratio of 1: 3: 5, then n = 7 and r = 3

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