Animated Solution for Mathematics - Functions: Let $a,b,c\in \mathbb{R}$. If $f(x)=a{x}^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy,\forall x,y\in \mathbb{R}$, then ${\sum }_{n=1}^{10}f(n)$ is equal to:

A Journey into Polynomials

Welcome, future engineer. Today, we are not just solving a math problem; we are embarking on a journey to decode the hidden structure of a function.

When you see a problem involving a functional equation like $f(x+y)=f(x)+f(y)+xy$, it is natural to feel a momentary spike of anxiety. It looks abstract, but remember, in the world of JEE Advanced, abstraction is just a veil waiting to be lifted. Let us pull back that veil together.

The Detective Work

We start with the general form of a quadratic function: $f(x)=a{x}^2+bx+c$. We are given a condition that the sum of the coefficients is three, so $a+b+c=3$.

In functional equations, the most powerful move is often the simplest one. We need to find the constant term $c$. As we discussed, $c$ is simply $f(0)$. So, let us test the equation at the origin by setting $x=0$ and $y=0$:

This simplifies to $f(0)=2f(0)$. Subtracting $f(0)$ from both sides, we find that $f(0)=0$.

Since $f(0)=c$, we have just unlocked our first secret: $c=0$. Our function is now significantly simpler: $f(x)=a{x}^2+bx$. We have already eliminated one variable.

The Algebraic Dance

Now, we need to find $a$. We take our simplified function $f(x)=a{x}^2+bx$ and substitute it back into the original functional equation. The left-hand side (LHS) becomes:

Expanding this, we get:

Now, look at the right-hand side (RHS). We have $f(x)+f(y)+xy$, which is:

When we equate the LHS and RHS, notice the magic of cancellation. The terms $a{x}^2$, $a{y}^2$, $bx$, and $by$ appear on both sides. They vanish, leaving us with a beautiful, clean relationship:

Since this must hold for all $x$ and $y$, we can equate the coefficients of the $xy$ term. Thus, $2a=1$, which gives us $a=\frac{1}{2}$.

The Final Piece

We have $a=\frac{1}{2}$ and $c=0$. We return to our initial condition: $a+b+c=3$. Substituting our known values:

We have fully reconstructed the function: $f(x)=\frac{1}{2}{x}^2+\frac{5}{2}x$. This is the heart of the problem.

The Summation

Finally, we must calculate the sum $S={\sum }_{n=1}^{10}f(n)$. Substituting our function, we get:

Using the linearity of summation, we split this into two parts:

We know our standard formulas. The sum of the first $N$ natural numbers is $\frac{N(N+1)}{2}$, and the sum of squares is $\frac{N(N+1)(2N+1)}{6}$. For $N=10$:

Plugging these back into our expression for $S$:

And there it is. The final answer is 330. You navigated the functional equation, performed the algebraic expansion, and executed the summation with precision. This is the mindset of a topper.