Question

If \(f(x)=ax^2+bx+c\) for some \(a,b,c\in \mathbb{R}\) with \(a+b+c=3\) and
\[ f(x+y)=f(x)+f(y)+xy \quad \forall x,y\in \mathbb{R}, \] then
\[ \sum_{n=1}^{10} f(n)= \]

• A. \(330\)
• B. \(255\)
• C. \(165\)
• D. \(190\)

Hint:

When a polynomial identity holds for all real \(x\) and \(y\), compare coefficients of like terms on both sides.

Answer 1

The Correct Option is A

Step 1: Write the given function.
Given,
\[ f(x)=ax^2+bx+c \]
Therefore,
\[ f(x+y)=a(x+y)^2+b(x+y)+c \]
\[ =ax^2+ay^2+2axy+bx+by+c \]

Step 2: Write \(f(x)+f(y)+xy\).
Now,
\[ f(x)+f(y)+xy \]
\[ =(ax^2+bx+c)+(ay^2+by+c)+xy \]
\[ =ax^2+ay^2+bx+by+2c+xy \]

Step 3: Compare both sides.
Since
\[ f(x+y)=f(x)+f(y)+xy \]
we compare the coefficients:
Coefficient of \(xy\):
\[ 2a=1 \]
\[ a=\frac12 \]
Constant term:
\[ c=2c \]
\[ c=0 \]

Step 4: Use the condition \(a+b+c=3\).
Given,
\[ a+b+c=3 \]
Substituting \(a=\frac12\) and \(c=0\),
\[ \frac12+b=3 \]
\[ b=\frac52 \]
Thus,
\[ f(x)=\frac12x^2+\frac52x \]

Step 5: Find \(\sum_{n=1}^{10}f(n)\).
\[ f(n)=\frac12n^2+\frac52n \]
So,
\[ \sum_{n=1}^{10}f(n)=\frac12\sum_{n=1}^{10}n^2+\frac52\sum_{n=1}^{10}n \]
Now,
\[ \sum_{n=1}^{10}n^2=385 \]
and
\[ \sum_{n=1}^{10}n=55 \]
Therefore,
\[ \sum_{n=1}^{10}f(n)=\frac12(385)+\frac52(55) \]
\[ =\frac{385}{2}+\frac{275}{2} \]
\[ =\frac{660}{2} \]
\[ =330 \]

Step 6: Final conclusion.
Hence,
\[ \boxed{330} \]