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 " sum_(n=1)^(10) f(n)` is equal to

Options

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

Explanation

To solve the problem step by step, we need to analyze the given conditions and derive the required values systematically.

### Step 1: Understand the function and conditions
We are given a quadratic function:
\[ f(x) = ax^2 + bx + c \]
with the conditions:
1. \( a + b + c = 3 \)
2. \( f(x+y) = f(x) + f(y) + xy \) for all \( x, y \in \mathbb{R} \)

### Step 2: Evaluate \( f(1) \)
Using the first condition, we can find \( f(1) \):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c = 3 \]

### Step 3: Use the second condition
Substituting \( y = 1 \) into the second condition:
\[ f(x+1) = f(x) + f(1) + x \]
This simplifies to:
\[ f(x+1) = f(x) + 3 + x \]

### Step 4: Find \( f(2) \)
Now, let’s find \( f(2) \) by substituting \( x = 1 \):
\[ f(2) = f(1) + 3 + 1 = 3 + 3 + 1 = 7 \]

### Step 5: Find \( f(3) \)
Next, we find \( f(3) \) by substituting \( x = 2 \):
\[ f(3) = f(2) + 3 + 2 = 7 + 3 + 2 = 12 \]

### Step 6: Identify a pattern
Continuing this process, we can find:
- For \( f(4) \):
\[ f(4) = f(3) + 3 + 3 = 12 + 3 + 3 = 18 \]
- For \( f(5) \):
\[ f(5) = f(4) + 3 + 4 = 18 + 3 + 4 = 25 \]
- For \( f(6) \):
\[ f(6) = f(5) + 3 + 5 = 25 + 3 + 5 = 33 \]
- For \( f(7) \):
\[ f(7) = f(6) + 3 + 6 = 33 + 3 + 6 = 42 \]
- For \( f(8) \):
\[ f(8) = f(7) + 3 + 7 = 42 + 3 + 7 = 52 \]
- For \( f(9) \):
\[ f(9) = f(8) + 3 + 8 = 52 + 3 + 8 = 63 \]
- For \( f(10) \):
\[ f(10) = f(9) + 3 + 9 = 63 + 3 + 9 = 75 \]

### Step 7: Calculate the sum \( \sum_{n=1}^{10} f(n) \)
Now, we sum the values from \( f(1) \) to \( f(10) \):
\[ \sum_{n=1}^{10} f(n) = f(1) + f(2) + f(3) + f(4) + f(5) + f(6) + f(7) + f(8) + f(9) + f(10) \]
\[ = 3 + 7 + 12 + 18 + 25 + 33 + 42 + 52 + 63 + 75 \]

Calculating this step-by-step:
- \( 3 + 7 = 10 \)
- \( 10 + 12 = 22 \)
- \( 22 + 18 = 40 \)
- \( 40 + 25 = 65 \)
- \( 65 + 33 = 98 \)
- \( 98 + 42 = 140 \)
- \( 140 + 52 = 192 \)
- \( 192 + 63 = 255 \)
- \( 255 + 75 = 330 \)

Thus, the final result is:
\[ \sum_{n=1}^{10} f(n) = 330 \]

### Final Answer:
The sum \( \sum_{n=1}^{10} f(n) \) is equal to \( 330 \).

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