Let `a,b,c in R.If f(x) = ax^2+bx +c ` is such that a +b+c =3 and `f(x+y)=f(x)+f(y)+xy,AA x,y in R, then Sigma_(n=1)^(10) f(n)` is equal to

Options

A. 255
B. 330
C. 165
D. 190

Explanation

To solve the problem, we need to find the value of \( \Sigma_{n=1}^{10} f(n) \) where \( f(x) = ax^2 + bx + c \) and it satisfies the conditions given in the problem.

### Step 1: Understand the function and conditions
We have:
1. \( f(x) = ax^2 + bx + c \)
2. \( a + b + c = 3 \)
3. \( f(x+y) = f(x) + f(y) + xy \)

### Step 2: Use the functional equation
To use the functional equation \( f(x+y) = f(x) + f(y) + xy \), we can substitute specific values for \( x \) and \( y \).

#### Substituting \( x = 1 \) and \( y = 1 \):
\[ f(1+1) = f(1) + f(1) + 1 \cdot 1 \]
This simplifies to:
\[ f(2) = 2f(1) + 1 \]

#### Now calculate \( f(1) \):
\[ f(1) = a(1^2) + b(1) + c = a + b + c = 3 \]
Thus, \( f(1) = 3 \).

#### Substitute \( f(1) \) into the equation for \( f(2) \):
\[ f(2) = 2(3) + 1 = 6 + 1 = 7 \]

### Step 3: Substitute \( x = 1 \) and \( y = 2 \):
\[ f(1+2) = f(1) + f(2) + 1 \cdot 2 \]
This simplifies to:
\[ f(3) = f(1) + f(2) + 2 \]
Substituting the values we found:
\[ f(3) = 3 + 7 + 2 = 12 \]

### Step 4: Set up equations for \( f(2) \) and \( f(3) \)
Now, we can express \( f(2) \) and \( f(3) \) in terms of \( a, b, c \):
- For \( f(2) \):
\[ f(2) = 4a + 2b + c = 7 \]
- For \( f(3) \):
\[ f(3) = 9a + 3b + c = 12 \]

### Step 5: Substitute \( c \) from the first condition
From \( a + b + c = 3 \), we can express \( c \):
\[ c = 3 - a - b \]

### Step 6: Substitute \( c \) into the equations
Substituting \( c \) into the equations for \( f(2) \) and \( f(3) \):
1. \( 4a + 2b + (3 - a - b) = 7 \)
\[ 3a + b + 3 = 7 \implies 3a + b = 4 \quad \text{(Equation 1)} \]
2. \( 9a + 3b + (3 - a - b) = 12 \)
\[ 8a + 2b + 3 = 12 \implies 8a + 2b = 9 \quad \text{(Equation 2)} \]

### Step 7: Solve the system of equations
From Equation 1:
\[ b = 4 - 3a \]
Substituting \( b \) into Equation 2:
\[ 8a + 2(4 - 3a) = 9 \]
\[ 8a + 8 - 6a = 9 \implies 2a = 1 \implies a = \frac{1}{2} \]
Substituting \( a \) back to find \( b \):
\[ b = 4 - 3\left(\frac{1}{2}\right) = 4 - \frac{3}{2} = \frac{5}{2} \]
Now substituting \( a \) and \( b \) back to find \( c \):
\[ c = 3 - \frac{1}{2} - \frac{5}{2} = 0 \]

### Step 8: Write the function
Thus, we have:
\[ f(x) = \frac{1}{2}x^2 + \frac{5}{2}x \]

### Step 9: Calculate \( \Sigma_{n=1}^{10} f(n) \)
We need to calculate:
\[ \Sigma_{n=1}^{10} f(n) = \Sigma_{n=1}^{10} \left( \frac{1}{2}n^2 + \frac{5}{2}n \right) \]
This can be separated into two sums:
\[ = \frac{1}{2} \Sigma_{n=1}^{10} n^2 + \frac{5}{2} \Sigma_{n=1}^{10} n \]

Using the formulas:
- \( \Sigma_{n=1}^{k} n = \frac{k(k+1)}{2} \)
- \( \Sigma_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \)

For \( k = 10 \):
\[ \Sigma_{n=1}^{10} n = \frac{10 \cdot 11}{2} = 55 \]
\[ \Sigma_{n=1}^{10} n^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]

Now substituting back:
\[ = \frac{1}{2} \cdot 385 + \frac{5}{2} \cdot 55 \]
\[ = \frac{385}{2} + \frac{275}{2} = \frac{660}{2} = 330 \]

### Final Answer
Thus, the value of \( \Sigma_{n=1}^{10} f(n) \) is \( \boxed{330} \).