Question

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

• A. $165$
• B. $190$
• C. $255$
• D. $336$

Answer 1

Answer: (A)

$165$

Given, $f\left(x\right)=a{x}^2+bx+c$,

$a+b+c=3.$

Also, $f(x+y)=f\left(x\right)+f\left(y\right)$

$\Rightarrow a{(x+y)}^2+b(x+y)+c=(a{x}^2+bx+c)+(a{y}^2+by+c)$

$\Rightarrow a(x^2+y^2+2xy)+b(x+y)+c=a(x^2+y^2)+b(x+y)+2c$

$\Rightarrow a(x^2+y^2)+2axy+b(x+y)+c=a(x^2+y^2)+b(x+y)+2c$

$\Rightarrow 2axy-c=0$

Now, Left-hand side gives $0$ for all real values of $x$ and $y$. Hence all coefficients must be $0$.

$\Rightarrow 2a=0\&c=0.$

$\Rightarrow a=0\&c=0.$

Since $a+b+c=3$,

$\Rightarrow 0+b+0=3$

$\Rightarrow b=3$

Therefore, $f\left(x\right)=3x$

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

$={\sum }_{n=1}^{10}3n$

$=3{\sum }_{n=1}^{10}n$

$=3.\frac{10(10+1)}{2}$ (Using ${\sum }_{r=1}^{n}r=\frac{n(n+1)}{2}$)

$=165.$

Hence, Option $\left(A\right)$ is correct.