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

Concept: Use the functional equation by substituting convenient values of \(x\) and \(y\) to determine \(f(n)\) and \(\alpha\).
Step 1: Given equation \[ f(x+y)=f(x)+2y^2+y+\alpha xy \]
Step 2: Find \(f(2)\) Put \(x=1, y=1\) \[ f(2)=f(1)+2(1)^2+1+\alpha(1)(1) \] \[ f(2)=2+2+1+\alpha \] \[ f(2)=5+\alpha \]
Step 3: Find \(f(3)\) Put \(x=2, y=1\) \[ f(3)=f(2)+2(1)^2+1+\alpha(2)(1) \] \[ f(3)=(5+\alpha)+3+2\alpha \] \[ f(3)=8+3\alpha \] Also put \(x=1, y=2\) \[ f(3)=f(1)+2(2)^2+2+\alpha(1)(2) \] \[ f(3)=2+8+2+2\alpha \] \[ f(3)=12+2\alpha \] Equating both expressions: \[ 8+3\alpha=12+2\alpha \] \[ \alpha=4 \]
Step 4: Find remaining values \[ f(2)=5+4=9 \] \[ f(3)=8+3(4)=20 \] Put \(x=3,y=1\) \[ f(4)=20+3+12=35 \] Put \(x=4,y=1\) \[ f(5)=35+3+16=54 \]
Step 5: Evaluate the sum \[ \sum_{i=1}^{5}(f(i)+\alpha) = \sum_{i=1}^{5}f(i)+5\alpha \] \[ =(-)+(2+9+20+35+54)+5(4) \] \[ =120+20 \] \[ =140 \]