If the constant term in the expansion of (1+2 x-3 x³)(3/2 x²-1/(3 x))⁹ is p, then 108 p is equal to ________.

Solution

Maths sir · 4 steps

Each of 1, 2x and -3x³ needs its own partner term from the bracket to make x⁰; find each one.

1. General term. In (3/2x²-1/3x)⁹, the (r+1)th term combines x²⁽⁹⁻ʳ⁾ and x⁻ʳ, so the power of x is 18-3r.

   Tᵣ₊₁ = C(9, r)(3/2)⁹⁻ʳ(-1/3)ʳ x¹⁸⁻³ʳ

2. Match each term. With 1 you need 18-3r=0, so r=6. With 2x you need 18-3r=-1, so r=19/3, which is not a whole number and gives nothing. With -3x³ you need 18-3r=-3, so r=7.
3. Compute both. For r=6: 84 × 27/8 × 1/729 = 7/18. For r=7: 36 × 9/4 × (-1/2187) = -1/27, and multiplying by -3 gives +1/9.
4. So. p = 7/18 + 1/9 = 9/18 = 1/2, and 108p = 54.

The common slip: 42. Taking only the 1 × T₇ term and forgetting the -3x³ contribution gives p = 7/18 and 42.

The idea to keep. When a polynomial multiplies a binomial power, pair each of its terms with the general term that cancels its power of x. Some pairings give a fractional r and contribute nothing.

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