In the expansion of (1+x)(1-x²)(1+3/x+3/x²+1/x³)⁵, x ≠ 0, the sum of the coefficients of x³ and x⁻¹³ is equal to __________.

Solution

Maths sir · 5 steps

Spot the perfect cube in the big bracket, and almost everything becomes one power of (1+x).

1. Spot the cube. The coefficients 1, 3, 3, 1 give 1+3/x+3/x²+1/x³ = (1+1/x)³ = (1+x)³/x³. Raising it to the fifth power gives (1+x)¹⁵/x¹⁵.
2. Combine everything. Factor 1-x² = (1-x)(1+x). With the other (1+x), the powers of (1+x) add up to 1+1+15 = 17.

   (1+x)(1-x²)(1+1/x)¹⁵ = (1-x)(1+x)¹⁷/x¹⁵

3. For x cubed. You need x¹⁸ in (1-x)(1+x)¹⁷. The highest power in (1+x)¹⁷ is x¹⁷, so only (-x) × x¹⁷ contributes, and the coefficient is -1.
4. For the negative power. For x⁻¹³ you need x² in (1-x)(1+x)¹⁷. That is C(17, 2) - C(17, 1) = 136 - 17 = 119.
5. So. The sum is 119 + (-1) = 118.

The common slip: 120. Taking the coefficient of x³ as +1 by dropping the minus sign from (1-x) gives 120, not 118.

The idea to keep. When a bracket has coefficients 1, 3, 3, 1 or 1, 2, 1, write it as a binomial power first. Then turn every wanted power of x into a power in the simplified numerator.

Answer from the NTA final answer key. Solution by PrepWiser.