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
$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy\,\,\,\forall x,y \in R$ then $\sum\limits_{n = 1}^8 {f\left( n \right)} $ is equal to
A.330
B.192
C.190
D.165

Answer 1

Hint: In this question there are two predefined functions. First$f\left( x \right) = a{x^2} + bx + c$ and $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy$ and value of $a + b + c = 3$. So make sure that we start to put the values of $x$ from $1$ to $8$. We will get a series of functions that will help to evaluate the next function value.

Complete step-by-step answer:
Given: $a,b,c \in R$, if $f\left( x \right) = a{x^2} + bx + c$ is such that $a + b + c = 3$ and $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy\,\,\,\forall x,y \in R$ then $\sum\limits_{n = 1}^8 {f\left( n \right)} $
Since, in the question $f\left( x \right)$ is defined as $f\left( x \right) = a{x^2} + bx + c$ so to solve it follow the function definition so,
$f\left( x \right) = a{x^2} + bx + c$ and $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy$
Now, $\sum\limits_{n = 1}^8 {f\left( n \right)} = f\left( 1 \right) + f\left( 2 \right) + f\left( 3 \right) + f\left( 4 \right) + f\left( 5 \right) + f\left( 6 \right) + f\left( 8 \right) + f\left( 8 \right)$
$\because f\left( 1 \right) = a{\left( 1 \right)^2} + b\left( 1 \right) + c = a + b + c = 3$
$f\left( 2 \right) = f\left( {1 + 1} \right) = f\left( 1 \right) + f\left( 1 \right) + 1 \times 1 = 3 + 3 + 1 = 7$ (Using $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy$)
$
  f\left( 3 \right) = f\left( {2 + 1} \right) = f\left( 2 \right) + f\left( 1 \right) + 2 \times 1 = 7 + 3 + 2 = 12 \\
  f\left( 4 \right) = f\left( {3 + 1} \right) = f\left( 3 \right) + f\left( 1 \right) + 3 \times 1 = 12 + 3 + 3 = 18 \\
  f\left( 5 \right) = f\left( {4 + 1} \right) = f\left( 4 \right) + f\left( 1 \right) + 4 \times 1 = 18 + 3 + 4 = 25 \\
  f\left( 6 \right) = f\left( {5 + 1} \right) = f\left( 5 \right) + f\left( 1 \right) + 5 \times 1 = 25 + 3 + 5 = 33 \\
  f\left( 7 \right) = f\left( {6 + 1} \right) = f\left( 6 \right) + f\left( 1 \right) + 6 \times 1 = 33 + 3 + 6 = 42 \\
  f\left( 8 \right) = f\left( {7 + 1} \right) = f\left( 7 \right) + f\left( 1 \right) + 7 \times 1 = 42 + 3 + 7 = 52 \\
$
Since, $\sum\limits_{n = 1}^8 {f\left( n \right)} = 3 + 7 + 12 + 18 + 25 + 33 + 42 + 52 = 192$
Hence, $\sum\limits_{n = 1}^8 {f\left( n \right)} = 192$
Hence option (b) is the correct answer.
So, the correct answer is “Option B”.

Note: In the question a student has to follow all the predefined functions and only have to put values according to the progressing way, and at last all the values for the final answer.