Let c, k ∈ R. If f(x) = (c + 1)x² + (1 - c²)x + 2k and f(x + y) = f(x) + f(y) - xy, for all x, y ∈ R, then the value of |2(f(1) + f(2) + f(3) + ...... + f(20))| is equal to ____________.

Solution

Maths sir · 4 steps

Put the quadratic into the equation and match the xy terms: that alone fixes c, and x = y = 0 fixes k.

1. Find k. Put x = y = 0: f(0) = 2f(0), so f(0) = 0. Since f(0) = 2k, we get k = 0.
2. Match the xy terms. Write f(x) = ax² + bx with a = c + 1. Then f(x+y) - f(x) - f(y) = 2axy, and this must equal -xy. So 2a = -1, which gives a = -1/2 and c = -3/2.
3. Find the function. The linear coefficient is b = 1 - c² = 1 - 9/4 = -5/4. So f(x) = -1/2x² - 5/4x.

   2Σₙ₌₁²⁰ f(n) = -Σₙ₌₁²⁰ n² - 5/2Σₙ₌₁²⁰ n = -2870 - 5/2× 210 = -3395

4. So. Using Σ n² = (20 × 21 × 41)/6 = 2870 and Σ n = 210, the sum times 2 is -3395. Its absolute value is 3395.

The common slip: 3185. Taking 2a = +1 instead of 2a = -1 gives c = -1/2, b = 3/4 and a final value of 3185, not 3395.

The idea to keep. When a functional equation is given for a known form like a polynomial, substitute the form and compare coefficients. Use x = y = 0 first to pin down the constant term.

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