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

Answer 1

The correct option is D

$330$

Explanation for correct answer:

Finding the $nth$term:

Given,

$$\begin{gathered} f\left(x\right)=a{x}^2+bx+c \\ f\left(1\right)=a+b+c=3 \end{gathered}$$

$f(x+y)=f\left(x\right)+f\left(y\right)+xy$

Substituting $x=1,y=1$

$$\begin{gathered} f(1+1)=f\left(1\right)+f\left(1\right)+1 \\ f\left(2\right)=2f\left(1\right)+1 \\ =2\times 3+1\left[\because f\left(1\right)=3\right] \\ =7 \end{gathered}$$

Similarly $x=2,y=1$

$$\begin{gathered} f(2+1)=f\left(2\right)+f\left(1\right)+\left(1\times 2\right) \\ f\left(3\right)=7+3+2\left[\because f\left(1\right)=3,f\left(2\right)=7\right] \\ =12 \end{gathered}$$

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

$$\begin{gathered} \sum _{n=1}^{10}f\left(n\right)=f\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(n\right) \\ =3+7+12+... \\ =S \end{gathered}$$

$S_n=3+7+12+...+t_n$

Again $S_n=3+7+12+...+t_{n-1}+t_n$

subtracting two-equation

$$\begin{gathered} S_n=3+7+12+...+t_{n-1}+t_n \\ -S_n=-3-7-12-...-t_{n-1}-t_n \\ 0=3+4+5+...-t_n \\ \Rightarrow t_n=3+4+5+...+n\text{terms} \end{gathered}$$

$$\begin{gathered} t_n=\frac{n(2\times 3+(n-1\left)1\right)}{2}\left[\because a_1+a_2+...+=\frac{n(2a+(n-1\left)d\right)}{2}\text{where}d=1\right] \\ =\frac{n(n+6-1)}{2} \\ =\frac{n(n+5)}{2} \end{gathered}$$

$$\begin{gathered} \sum _{n=1}^{10}{t}_n=\sum _{n=1}^{10}\frac{n^2+5n}{2} \\ =\sum _{n=1}^{10}\frac{n^2}{2}+\frac{5n}{2} \\ =\frac{1}{2}\left\{\sum _{n=1}^{10}{n}^2+5\underset{n=1}{\overset{10}{\sum n}}\right\} \\ =\frac{1}{2}\left\{\frac{10\times 11\times 21}{6}+5\frac{10\times 11}{2}\right\} \\ =\frac{1}{2}\frac{10\times 11}{2}\left\{\frac{21}{3}+5\right\} \\ =\frac{5\times 11}{2}\left\{\frac{21+15}{3}\right\} \\ =\frac{55}{2}\left\{\frac{36}{3}\right\} \\ =55\times 6 \\ =330 \end{gathered}$$

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

Hence, option (D) is the correct answer.