Question

If \(f(x)\) satisfies the functional equation \[ f(x+y)=f(x)+2y^2+y+\alpha xy \] where \(x,y\) are whole numbers, such that \(f(0)=-1\) and \(f(1)=2\), then the value of \[ \sum_{i=1}^{5}\big(f(i)+\alpha\big) \] is

• A. \(130\)
• B. \(145\)
• C. \(120\)
• D. \(140\)

Hint:

For functional equations, substitute small integer values first to determine unknown constants.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to find the value of the constant \( \alpha \) and the general form of the function \( f(x) \) using the given functional equation and initial conditions.
Step 2: Key Formula or Approach:
1. Set \( x=0 \) to find a simpler form for \( f(y) \). 2. Use \( f(1) = 2 \) to find \( \alpha \).
Step 3: Detailed Explanation:
1. Let \( x=0 \): \( f(0 + y) = f(0) + 2y^2 + y + \alpha(0)y \) \[ f(y) = -1 + 2y^2 + y = 2y^2 + y - 1 \] 2. Check with \( f(1) = 2 \): \[ f(1) = 2(1)^2 + 1 - 1 = 2 \] (This matches, but we need to find \( \alpha \) from the interaction term). 3. Use \( f(x+y) \) again: \( f(1+1) = f(1) + 2(1)^2 + 1 + \alpha(1)(1) \implies f(2) = 2 + 2 + 1 + \alpha = 5 + \alpha \). From the formula \( f(y) = 2y^2 + y - 1 \), \( f(2) = 2(4) + 2 - 1 = 9 \). So, \( 5 + \alpha = 9 \implies \alpha = 4 \). 4. Calculate the sum \( \sum_{i=1}^5 (2i^2 + i - 1 + 4) = \sum_{i=1}^5 (2i^2 + i + 3) \): \[ 2\sum i^2 = 2(55) = 110 \] \[ \sum i = 15 \] \[ \sum 3 = 15 \] Total sum = \( 110 + 15 + 15 = 140 \).
Step 4: Final Answer:
The value of the sum is 140.