Let a, b, c ∈ R . If f(x) = a x ^ 2 + bx + c is such that a + b + c...

Given, $f\left(x + y\right) = f\left(x\right) + f\left(y\right) + xy \forall x, y \in R$

Put $y = 1$, we get

$⇒ f\left(x + 1\right) = f\left(x\right) + f\left(1\right) + x$

$⇒ f\left(x + 1\right) - f\left(x\right) = 3 + x$

On taking Sigma on both sides

$⇒ \mathop{∑}_{x = 1}^{n - 1}\left(f\left(x + 1\right) - f\left(x\right)\right) = \mathop{∑}_{x = 1}^{n - 1}\left(3 + x\right)$

$⇒ \left(f\left(2\right) - f\left(1\right)\right) + \left(f\left(3\right) - f\left(2\right)\right) + \left(f\left(4\right) - f\left(3\right)\right) + \left(f\left(5\right) - f\left(4\right)\right) + . . . . . . + \left(f\left(n\right) - f\left(n - 1\right)\right)$

$= 3\left(n - 1\right) + \dfrac{\left(n - 1\right)n}{2}$

$⇒ f\left(n\right) - f\left(1\right) = 3\left(n - 1\right) + \dfrac{\left(n - 1\right)n}{2}$

$⇒ f\left(n\right) - 3 = 3\left(n - 1\right) + \dfrac{\left(n - 1\right)n}{2}$

$⇒ f\left(n\right) = \dfrac{{n}^{2}}{2} + \dfrac{5n}{2}$

Again taking Sigma on both the sides

$⇒ \mathop{∑}_{n = 1}^{10}f\left(n\right) = \mathop{∑}_{n = 1}^{10}\left(\dfrac{{n}^{2}}{2} + \dfrac{5n}{2}\right)$

$⇒ \mathop{∑}_{n = 1}^{10}f\left(n\right) = \dfrac{1}{2} . \dfrac{10 . 11 . 21}{6} + \dfrac{5}{2} . \dfrac{10 . 11}{2}$

$⇒ \mathop{∑}_{n = 1}^{10}f\left(n\right) = 330$.