Question

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

• A. 330
• B. 165
• C. 190
• D. 255

Answer 1

The correct option is A 330

We have, $f\left(x\right)=a{x}^2+bx+c$
Now, $f(x+y)=f\left(x\right)+f\left(y\right)+xy$
Put $y=0\Rightarrow f\left(x\right)=f\left(x\right)+f\left(0\right)+0$
$\Rightarrow f\left(0\right)=0$
$\Rightarrow c=0$
Again, put $y=-x$
$\therefore f\left(0\right)=f\left(x\right)+f(-x)-x^2$
$\Rightarrow 0=a{x}^2+bx+a{x}^2-bx-x^2\Rightarrow 2a{x}^2-x^2=0\Rightarrow a=\frac{1}{2}$
Also, $a+b+c=3$
$\Rightarrow \frac{1}{2}+b+0=3\Rightarrow b=\frac{5}{2}\therefore f\left(x\right)=\frac{x^2+5x}{2}\text{Now},f\left(n\right)=\frac{n^2+5n}{2}=\frac{1}{2}{n}^2+\frac{5}{2}n$
$\therefore {\sum }_{n=1}^{10}f\left(n\right)=\frac{1}{2}{\sum }_{n=1}^{10}{n}^2+\frac{5}{2}{\sum }_{n=1}^{10}n=\frac{1}{2}\cdot \frac{10\times 11\times 21}{6}+\frac{5}{2}\cdot \frac{10\times 11}{2}=\frac{385}{2}+\frac{275}{2}=\frac{660}{2}=330$