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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 ...

Let a,b,c∈Ra,b,c \in R, if f(x)=ax2+bx+cf\left( x \right) = a{x^2} + bx + c is such that a+b+c=3a + b + c = 3 and
f(x+y)=f(x)+f(y)+xy   ∀x,y∈Rf\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy\,\,\,\forall x,y \in R then ∑n=18f(n)\sum\limits_{n = 1}^8 {f\left( n \right)} is equal to
A.330
B.192
C.190
D.165

Explanation

Solution

Hint : In this question there are two predefined functions. Firstf(x)=ax2+bx+cf\left( x \right) = a{x^2} + bx + c and f(x+y)=f(x)+f(y)+xyf\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy and value of a+b+c=3a + b + c = 3. So make sure that we start to put the values of xx from 11 to 88. 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∈Ra,b,c \in R, if f(x)=ax2+bx+cf\left( x \right) = a{x^2} + bx + c is such that a+b+c=3a + b + c = 3 and f(x+y)=f(x)+f(y)+xy   ∀x,y∈Rf\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy\,\,\,\forall x,y \in R then ∑n=18f(n)\sum\limits_{n = 1}^8 {f\left( n \right)}
Since, in the question f(x)f\left( x \right) is defined as f(x)=ax2+bx+cf\left( x \right) = a{x^2} + bx + c so to solve it follow the function definition so,
f(x)=ax2+bx+cf\left( x \right) = a{x^2} + bx + c and f(x+y)=f(x)+f(y)+xyf\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy
Now, ∑n=18f(n)=f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(8)+f(8)\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)
∵f(1)=a(1)2+b(1)+c=a+b+c=3\because f\left( 1 \right) = a{\left( 1 \right)^2} + b\left( 1 \right) + c = a + b + c = 3
f(2)=f(1+1)=f(1)+f(1)+1×1=3+3+1=7f\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(x+y)=f(x)+f(y)+xyf\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy)
f(3)=f(2+1)=f(2)+f(1)+2×1=7+3+2=12 f(4)=f(3+1)=f(3)+f(1)+3×1=12+3+3=18 f(5)=f(4+1)=f(4)+f(1)+4×1=18+3+4=25 f(6)=f(5+1)=f(5)+f(1)+5×1=25+3+5=33 f(7)=f(6+1)=f(6)+f(1)+6×1=33+3+6=42 f(8)=f(7+1)=f(7)+f(1)+7×1=42+3+7=52  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, ∑n=18f(n)=3+7+12+18+25+33+42+52=192\sum\limits_{n = 1}^8 {f\left( n \right)} = 3 + 7 + 12 + 18 + 25 + 33 + 42 + 52 = 192
Hence, ∑n=18f(n)=192\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.