Question

Find the coefficient of:

(i) x^m in the expansion of (x + 1/x)ⁿ

(ii) x in the expansion of (1 – 2x³ + 3x⁵) (1 + 1/x)⁸

(iii) a⁵b⁷ in the expansion of (a – 2b)¹²

(iv) x in the expansion of (1 – 3x + 7x²) (1 – x)¹⁶

Answer 1

(i) x^m in the expansion of (x + 1/x)ⁿ

Given as

(x + 1/x)ⁿ

If x^m occurs at the (r + 1)^th term in the given expression.
Now, we have:

T_r+1 = ⁿC_r x^n-r a^r

(ii) x in the expansion of (1 – 2x³ + 3x⁵) (1 + 1/x)⁸

Given as

(1 – 2x³ + 3x⁵) (1 + 1/x)⁸

If x occurs at the (r + 1)^th term in the given expression.
Now, we have:

(1 – 2x³ + 3x⁵) (1 + 1/x)⁸ = (1 – 2x³ + 3x⁵) (⁸C₀ + ⁸C₁ (1/x) + ⁸C₂ (1/x)² + ⁸C₃ (1/x)³ + ⁸C₄ (1/x)⁴ + ⁸C₅ (1/x)⁵ + ⁸C₆ (1/x)⁶ + ⁸C₇ (1/x)⁷ + ⁸C₈ (1/x)⁸)

Therefore, ‘x’ occurs in the above expression at -2x³.⁸C₂ (1/x²) + 3x⁵.⁸C₄ (1/x⁴)

∴ Coefficient of x = -2 (8!/(2!6!)) + 3 (8!/(4! 4!))

= -56 + 210

= 154

(iii) a⁵b⁷ in the expansion of (a – 2b)¹²

Given as

(a – 2b)¹²

If a⁵b⁷ occurs at the (r + 1)^th term in the given expression.
Now, we have:

T_r+1 = ⁿC_r x^n-r a^r

(iv) x in the expansion of (1 – 3x + 7x²) (1 – x)¹⁶

Given as

(1 – 3x + 7x²) (1 – x)¹⁶

If x occurs at the (r + 1)^th term in the given expression.
Now, we have:

(1 – 3x + 7x²) (1 – x)¹⁶ = (1 – 3x + 7x²) (¹⁶C₀ + ¹⁶C₁ (-x) + ¹⁶C₂ (-x)² + ¹⁶C₃ (-x)³ + ¹⁶C₄ (-x)⁴ + ¹⁶C₅ (-x)⁵ + ¹⁶C₆ (-x)⁶ + ¹⁶C₇ (-x)⁷ + ¹⁶C₈ (-x)⁸ + ¹⁶C₉ (-x)⁹ + ¹⁶C₁₀ (-x)¹⁰ + ¹⁶C₁₁ (-x)¹¹ + ¹⁶C₁₂ (-x)¹² + ¹⁶C₁₃ (-x)¹³ + ¹⁶C₁₄ (-x)¹⁴ + ¹⁶C₁₅ (-x)¹⁵ + ¹⁶C₁₆ (-x)¹⁶)

Therefore, ‘x’ occurs in the above expression at ¹⁶C₁ (-x) – 3x¹⁶C₀

∴ Coefficient of x = -(16!/(1! 15!)) – 3(16!/(0! 16!))

= -16 – 3

= -19