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

Answer 1

Answer: (B)

$330$

$f\left(1\right)=a+b+c\Rightarrow f\left(1\right)=3f\left(2\right)=f(1+1)=f\left(1\right)+f\left(1\right)+1\times 1=3+3+1=7f\left(3\right)=f(2+1)=f\left(2\right)+f\left(1\right)+2\times 1=7+3+2=12$

Similarly, $f\left(4\right)=18f\left(5\right)=25f\left(6\right)=33f\left(7\right)=42f\left(8\right)=52f\left(9\right)=63f\left(10\right)=75$

and

$f\left(1\right)+f\left(2\right)+f\left(3\right)+f\left(4\right)+f\left(5\right)+f\left(6\right)+f\left(7\right)+f\left(8\right)+f\left(9\right)+f\left(10\right)=330$

Therefore, correct answer is $330$.