Question

Let \( f : \mathbb{N} \to \mathbb{N} \) be a function such that \( f(x + y) = f(x) + f(y) + xy \) for every \( x, y \in \mathbb{N} \). If \( f(1) = 2 \), then \( \sum\limits_{k=0}^{10} f(k) \) is equal to:

• A. \(1650\)
• B. \(275\)
• C. \(550\)
• D. \(1025\)

Hint:

When given a functional equation, test it by plugging small values to detect a pattern. Recurrence or recursive computation often helps in such functional equations.

Answer 1

The Correct Option is B

Step 1: Given functional equation:
\[ f(x+y) = f(x) + f(y) + xy \quad \text{and} \quad f(1) = 2 \]
Step 2: Find \( f(0) \):
Using the functional equation:
\[ f(x+0) = f(x) + f(0) + 0 \cdot x \Rightarrow f(x) = f(x) + f(0) \Rightarrow f(0) = 0 \]
Step 3: Compute values recursively:
\( f(1) = 2 \) (Given)
\( f(2) = f(1+1) = f(1) + f(1) + 1 \cdot 1 = 2 + 2 + 1 = 5 \)
\( f(3) = f(2+1) = f(2) + f(1) + 2 \cdot 1 = 5 + 2 + 2 = 9 \)
\( f(4) = f(3+1) = f(3) + f(1) + 3 \cdot 1 = 9 + 2 + 3 = 14 \)
\( f(5) = f(4+1) = f(4) + f(1) + 4 \cdot 1 = 14 + 2 + 4 = 20 \)
\( f(6) = f(5+1) = f(5) + f(1) + 5 \cdot 1 = 20 + 2 + 5 = 27 \)
\( f(7) = f(6+1) = f(6) + f(1) + 6 \cdot 1 = 27 + 2 + 6 = 35 \)
\( f(8) = f(7+1) = f(7) + f(1) + 7 \cdot 1 = 35 + 2 + 7 = 44 \)
\( f(9) = f(8+1) = f(8) + f(1) + 8 \cdot 1 = 44 + 2 + 8 = 54 \)
\( f(10) = f(9+1) = f(9) + f(1) + 9 \cdot 1 = 54 + 2 + 9 = 65 \)

Step 4: Compute the total sum:
\[ \sum_{k=0}^{10} f(k) = 0 + 2 + 5 + 9 + 14 + 20 + 27 + 35 + 44 + 54 + 65 = 275 \]
Final Answer: 275