Question

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

• A. $330$
• B. $165$
• C. $190$
• D. $255$

Answer 1

The correct option is A $330$

$f\left(x\right)=a{x}^2+bx+c$
and $f(x+y)=f\left(x\right)+f\left(y\right)+xy$
$\Rightarrow a(x+y{)}^2+b(x+y)+c=(a{x}^2+bx+c)+(a{y}^2+by+c)+xy$
$\Rightarrow 2axy=xy+c$
$\Rightarrow a=\frac{1}{2}$ and $c=0$
Also, $a+b+c=3\Rightarrow b=\frac{5}{2}$
$\therefore f\left(x\right)=\frac{1}{2}{x}^2+\frac{5}{2}x$

Then $\sum _{n=1}^{10}f\left(n\right)=\frac{1}{2}\sum _{n=1}^{10}{n}^2+\frac{5}{2}\sum _{n=1}^{10}n$
$=\frac{1}{2}\times \frac{10\times 11\times 21}{6}+\frac{5}{2}\times \frac{10\times 11}{2}$
$=330$