Let the sum of the coefficients of the first three terms in the expansion of ( x - 3/x² )ⁿ,x ≠ 0. n ∈ ℕ, be 376. Then the coefficient of x⁴ is __________.

Solution

Maths sir · 4 steps

Use the 376 condition to find n first; then one general term gives you the x^4 coefficient.

1. First three coefficients. The general term is Tᵣ₊₁ = C(n, r) xⁿ⁻ʳ (-3/x²)ʳ = C(n, r)(-3)ʳ xⁿ⁻³ʳ. For r = 0, 1, 2 the coefficients are 1, -3n and 9C(n, 2).

   1 - 3n + 9n(n-1)/2 = 376

2. Solve for n. Multiply by 2 to get 9n² - 15n - 750 = 0, that is 3n² - 5n - 250 = 0. Then n = (5 ± √(25 + 3000))/6 = (5 ± 55)/6, and since n is a natural number, n = 10.
3. Pick the term. With n = 10 the power of x is 10 - 3r. Setting 10 - 3r = 4 gives r = 2.
4. So. The coefficient of x⁴ is C(10, 2)(-3)² = 45 × 9 = 405.

The common slip: 45. Dropping the factor (-3)² and taking only C(10, 2) gives 45, not 405. The constant in the second term is part of the coefficient.

The idea to keep. In (x + a/xᵏ)ⁿ, write the general term with its full power of x first. The coefficient carries aʳ with its sign, so keep (-3)ʳ intact.

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