O-Level Binomial expansions: worked solution

Question

Using that k, show there is no constant term in (1+x⁴)(x−k/x³)^8.

The first bracket (1 + x⁴) contributes either 1 or x⁴, so there are exactly two ways of producing a constant term (a term in x⁰) in the product: 1 multiplied by the constant term of the expansion of (x − k/x³)⁸, and x⁴ multiplied by the term in x⁻⁴ of that same expansion. Step 1: write down the general term of (x − k/x³)⁸. Using the binomial theorem with first part x and second part (−k/x³): general term = C(8, r) × x^(8−r) × (−k/x³)^r = C(8, r) × (−k)^r × x^(8−r) × x^(−3r) = C(8, r) × (−k)^r × x^(8−4r). Step 2: find the constant term of the bracket. A constant term needs the power of x to be zero: 8 − 4r = 0, so 4r = 8 and r = 2. With r = 2: C(8, 2)(−k)² = 28 × k² = 28k² (the square removes the minus sign). Step 3: find the term in x⁻⁴. 8 − 4r = −4, so 4r = 12 and r = 3. With r = 3: C(8, 3)(−k)³ = 56 × (−k³) = −56k³, giving the term −56k³x⁻⁴. Step 4: put in the value of k found in part (a), k = 1/2. Constant term: 28k² = 28 × (1/2)² = 28 × 1/4 = 7. Term in x⁻⁴: −56k³ = −56 × (1/2)³ = −56 × 1/8 = −7, i.e. −7x⁻⁴. Step 5: collect the two contributions to the constant term of the whole product. from 1 × 7 = 7 from x⁴ × (−7x⁻⁴) = −7x⁰ = −7 Total constant term = 7 + (−7) = 0. The two contributions cancel exactly, so (1 + x⁴)(x − k/x³)⁸ has no constant term, as required.