1. Given

We have $f(x)=ax^2+bx+c$ and $a+b+c=3$ with the functional equation $f(x+y)=f(x)+f(y)+xy \,\,\forall x,y\in\mathbb R.$

We need to find $\sum_{n=1}^{10} f(n).$

2. Expand both sides of the functional equation

Since $f(x)=ax^2+bx+c,$ we get $f(x+y)=a(x+y)^2+b(x+y)+c.$ Expanding, $f(x+y)=ax^2+2axy+ay^2+bx+by+c.$

Now, $f(x)+f(y)+xy=(ax^2+bx+c)+(ay^2+by+c)+xy$ $=ax^2+ay^2+bx+by+2c+xy.$

Since these are equal for all $x,y$, compare coefficients:

• Coefficient of $xy$: $2a=1 \implies a=\frac12$
• Constant term: $c=2c \implies c=0$

Now use $a+b+c=3$ so $\frac12+b+0=3$ $b=\frac52.$

Hence, $f(x)=\frac12x^2+\frac52x.$

3. Compute $f(n)$ and the required sum

For integer $n$, $f(n)=\frac12n^2+\frac52n.$ Therefore, $\sum_{n=1}^{10} f(n)=\frac12\sum_{n=1}^{10} n^2+\frac52\sum_{n=1}^{10} n.$

Use standard sums: $\sum_{n=1}^{10} n=\frac{10\cdot 11}{2}=55,$ $\sum_{n=1}^{10} n^2=\frac{10\cdot 11\cdot 21}{6}=385.$

So, $\sum_{n=1}^{10} f(n)=\frac12(385)+\frac52(55)$ $=192.5+137.5=330.$

4. Check options

The value is $330,$ which matches Option D.

5. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.