Let a ,   b ,   c ∈ R . If f x = a x 2 + b x + c is such that a + b + c = 3 and f x + y = f…

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

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

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

On taking Sigma on both sides

$\Rightarrow \sum _{x=1}^{n-1}\left(f\left(x+1\right)-f\left(x\right)\right)=\sum _{x=1}^{n-1}\left(3+x\right)$

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

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

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

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

$\Rightarrow f\left(n\right)=\frac{n^2}{2}+\frac{5n}{2}$

Again taking Sigma on both the sides 

$\Rightarrow \sum _{n=1}^{10}f\left(n\right)=\sum _{n=1}^{10}\left(\frac{n^2}{2}+\frac{5n}{2}\right)$

$\Rightarrow \sum _{n=1}^{10}f\left(n\right)=\frac{1}{2}.\frac{10.11.21}{6}+\frac{5}{2}.\frac{10.11}{2}$

$\Rightarrow \sum _{n=1}^{10}f\left(n\right)=330$.