Find the 13ᵗʰ term in the expansion of (9x - 1/3√x)¹⁸, x ≠ 0

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.

Thus, the 13ᵗʰ term in the expansion of (9x - 1/3√x)¹⁸ is

T₁₃ = T₁₂ ₊ ₁

= ¹⁸C₁₂ (9x)¹⁸ ⁻ ¹² (- 1/3√x)¹²

= [18! / (12!)(6!)] x 9⁶ x (x)⁶ (- 1/3)¹² (- 1/√x)¹²

= [18 x 17 x 16 x 15 x 14 x 13 x (12!)] / [(12!) x 6 x 5 x 4 3 x 2] x 3¹² x (1/3¹²) x (x)⁶ x (1/x)⁶ [∵ 9⁶ = (3²)⁶ = 3¹²]

= 18564

NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.2 Question 6

Find the 13ᵗʰ term in the expansion of (9x - 1/3√x)¹⁸, x ≠ 0

Summary:

Using binomial theorem, 13ᵗʰ term in the expansion of (9x - 1/3√x)¹⁸, x ≠ 0 is to be found. We have found that it is equal to 18564