Let (1+x+x²)¹⁰=a₀+a₁ x+a₂ x²+…+a₂₀ x²⁰. If (a₁+a₃+a₅+…+a₁₉)-11 a₂=121 k, then k is equal to_________ .

Solution

Maths sir · 4 steps

Put x = 1 and x = -1 to get the odd coefficients, then count the ways to build x² by hand.

1. Odd coefficients. Let f(x) = (1+x+x²)¹⁰. Then f(1) - f(-1) cancels every even-power coefficient and doubles every odd one. Here f(1) = 3¹⁰ = 59049 and f(-1) = 1¹⁰ = 1.

   a₁ + a₃ + … + a₁₉ = (f(1) - f(-1))/2 = (59049 - 1)/2 = 29524

2. Find a2. There are two ways to get x² from the 10 brackets. You can take x² from one bracket, in C(10, 1) = 10 ways, or x from two brackets, in C(10, 2) = 45 ways. So a₂ = 55 and 11a₂ = 605.
3. Divide by 121. 29524 - 605 = 28919, and 28919 = 121 × 239.
4. So. k = 239.

The common slip: 483. Forgetting to halve f(1) - f(-1) makes the odd sum 59048, and then (59048 - 605)/121 gives 483.

The idea to keep. To sum the odd-power coefficients, use (f(1) - f(-1))/2. For a small coefficient of a trinomial power, list every way to build that power from the brackets.

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