Solution: Assume $f$ is quadratic. Let $f(x) = px^2 + qx + r$. Substitute into the equation: $p(a + b)^2 + q(a + b) + r = pa^2 + qa + r + pb^2 + qb + r + ab$. Expand and equate coefficients: $p(a^2 + 2ab + b^2) + q(a + b) + r = pa^2 + pb^2 + q(a + b) + 2r + ab$. Simplify: $2pab = ab + 2r$. For this to hold for all $a, b$, we require $2p = 1$ and $2r = 0$, so $p = rac{1}{2}$, $r = 0$. The linear term $q$ cancels out, so $f(x) = rac{1}{2}x^2 + qx$. Verifying, $f(a + b) = rac{1}{2}(a + b)^2 + q(a + b) = rac{1}{2}a^2 + ab + rac{1}{2}b^2 + q(a + b)$, and $f(a) + f(b) + ab = rac{1}{2}a^2 + qa + rac{1}{2}b^2 + qb + ab$. The results match. Thus, all solutions are $f(x) = oxed{\dfrac{1}{2}x^2 + cx}$ for some constant $c \in \mathbb{R}$.Question: A conservation educator observes that the population of a rare bird species increases by a periodic pattern modeled by $ P(n) = n^2 + 3n + 5 $, where $ n $ is the year modulo 10. What is the remainder when $ P(1) + P(2) + \dots + P(10) $ is divided by 7?

Sat, 02 May 2026 20:46:18 +0000

Try to find the remainder when $ S = P(1) + P(2) + \dots + P(10) $ is divided by 7, where $ P(n) = n^2 + 3n + 5 $. First, compute $ S = \sum_{n=1}^{10} (n^2 + 3n + 5) $. Break into parts: $$
S = \sum_{n=1}^{10} n^2 + 3 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 5
$$ Use standard summation formulas: