The absolute difference of the coefficients of x¹⁰ and x⁷ in the expansion of (2 x²+1/(2 x))¹¹ is equal to :

Solution

Maths sir · 4 steps

Write the general term once, collect the powers of x and of 2, and each coefficient falls out.

1. General term. Take (r+1)-th term Tᵣ₊₁ = C(11, r)(2x²)¹¹⁻ʳ(1/2x)ʳ. Collect the powers of 2 and of x separately.

   Tᵣ₊₁ = C(11, r) 2¹¹⁻²ʳ x²²⁻³ʳ

2. Find each r. For x¹⁰, 22-3r=10 gives r=4, so the coefficient is C(11, 4)× 2³ = 330 × 8 = 2640. For x⁷, 22-3r=7 gives r=5, so the coefficient is C(11, 5)× 2¹ = 462 × 2 = 924.
3. Match an option. The difference is 2640 - 924 = 1716. Now 12³ - 12 = 1728 - 12 = 1716. You can also see it as 11 × 12 × 13 = 1716.
4. So. The absolute difference is 1716 = 12³ - 12, which is option (c).

The idea to keep. In any coefficient problem, first simplify the general term to the form (number) × xᵖᵒʷᵉʳ. Then set the power equal to what you need and solve for r.

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