Let a, b, c ∈ R. If f(x) = ax² + bx + c is such that a + b + c = 3 and f(x+y) = f(x) + f(y) + xy, ∀ x, y ∈ R, then ∑ₙ₌₁¹⁰ f(n) is equal to:

Question

Accepted Answer

330. From f(x+y)=f(x)+f(y)+xy we get 2a = 1 so a=1/2 and c=0; then a+b+c=3 gives b=5/2, so f(n)=n^2/2 + 5n/2. Sum for n=1..10 = (1/2)(385) + (5/2)(55) = 192.5 + 137.5 = 330.

Let a, b, c ∈ R. If f(x) = ax² + bx + c is such that a + b + c = 3 and f(x+y) = f(x) + f(y) + xy, ∀ x, y ∈ R, then ∑ₙ₌₁¹⁰ f(n) is equal to:

Solution

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