Understanding the Problem

We are given a quadratic function $f\left(x\right)=a{x}^2+bx+c$ with real coefficients, satisfying two conditions:

1. $a+b+c=3$
2. $f(x+y)=f\left(x\right)+f\left(y\right)+xy$ for all real $x$, $y$

We need to find the value of the sum $\sum _{n=1}^{10}f\left(n\right)$.

Step 1: Determine the Coefficients a, b, c

Let's use the given functional equation to find relationships between the coefficients.

Substitute $x=0$, $y=0$ into the equation:

$f(0+0)=f\left(0\right)+f\left(0\right)+\left(0\right)\left(0\right)$

$f\left(0\right)=2f\left(0\right)$

This implies $f\left(0\right)=0$.

But from the definition, $f\left(0\right)=a(0{)}^2+b(0)+c=c$, so $c=0$.

Now, substitute $y=-x$ into the equation:

$f(x+(-x\left)\right)=f\left(x\right)+f(-x)+x(-x)$

$f\left(0\right)=f\left(x\right)+f(-x)-x^2$

Since $f\left(0\right)=0$, we have:

$0=f\left(x\right)+f(-x)-x^2$

$f\left(x\right)+f(-x)=x^2$ (Equation A)

Now, let's compute $f\left(x\right)+f(-x)$ directly from the quadratic form (with $c=0$):

$f\left(x\right)+f(-x)=[a{x}^2+bx]+[a{x}^2-bx]=2a{x}^2$

Set this equal to Equation A:

$2a{x}^2=x^2$

This must hold for all $x$, so $2a=1$, which gives $a=\frac{1}{2}$.

Now, use the first condition $a+b+c=3$. We have $a=\frac{1}{2}$, $c=0$, so:

$\frac{1}{2}+b+0=3$

$b=3-\frac{1}{2}=\frac{5}{2}$

Therefore, the function is:

$f\left(x\right)=\frac{1}{2}{x}^2+\frac{5}{2}x$

Step 2: Compute the Sum from n=1 to 10

We need to find $S=\sum _{n=1}^{10}f\left(n\right)=\sum _{n=1}^{10}(\frac{1}{2}{n}^2+\frac{5}{2}n)$

This can be separated into two sums:

$S=\frac{1}{2}\sum _{n=1}^{10}{n}^2+\frac{5}{2}\sum _{n=1}^{10}n$

We use the standard summation formulas:

$\sum _{n=1}^{k}n=\frac{k(k+1)}{2}$

$\sum _{n=1}^k{n}^2=\frac{k(k+1)(2k+1)}{6}$

For $k=10$:

$\sum _{n=1}^{10}n=\frac{10\times 11}{2}=55$

$\sum _{n=1}^{10}{n}^2=\frac{10\times 11\times 21}{6}=\frac{2310}{6}=385$

Now substitute back:

$S=\frac{1}{2}\left(385\right)+\frac{5}{2}\left(55\right)=\frac{385}{2}+\frac{275}{2}=\frac{660}{2}=330$

Final Answer

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

Related Topics & Formulae

Quadratic Functions

A quadratic function is of the form $f\left(x\right)=a{x}^2+bx+c$, where $a\ne 0$. Its graph is a parabola.

Functional Equations

Equations where the unknown is a function. Solving often involves substituting specific values (like 0, 1, -x) to find relationships between the function's parameters or form.

Summation Formulas

Key formulas for summing sequences:

• Sum of first n natural numbers: $\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$
• Sum of squares of first n natural numbers: $\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$
• Sum of cubes: $\sum _{k=1}^n{k}^3={\left(\frac{n(n+1)}{2}\right)}^2$