Question

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

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

Answer 1

The correct option is A $330$

Let $a,b,cϵ$ R

$f\left(1\right)=a+b+c=3$

$f\left(2\right)=f(1+1)=f\left(1\right)+f\left(1\right)+1$ ............. $\because f(x+y)=f\left(x\right)+f\left(y\right)+xy$

$⟹f\left(2\right)=2f\left(1\right)+1$

$f\left(3\right)=f(2+1)=f\left(2\right)+f\left(1\right)+2$

$=2f\left(1\right)+1+f\left(1\right)+2$

$⟹f\left(3\right)=3f\left(1\right)+3$

$f\left(4\right)=f(3+1)=f\left(3\right)+f\left(1\right)+3$

$=3f\left(1\right)+3+f\left(1\right)+3$

$⟹f\left(4\right)=4f\left(1\right)+6$

$\sum _{n=1}^{10}f\left(n\right)=f\left(1\right)+f\left(2\right)+f\left(3\right)+--------+f\left(10\right)$

$=f\left(1\right)+2f\left(1\right)+1+3f\left(1\right)+3+4f\left(1\right)+6+5f\left(1\right)+10--------$

$=f\left(1\right)[1+2+3+4+-----+10]+(1+3+6+10+15+21+28+36+45)$

$=f\left(1\right)\left(55\right)+\left(165\right)$

$=3\times 55+165=330$

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

Hence, option A is correct.