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Functional Equation Sum Σ(α+f(n)) = 140 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let for some α∈R\alpha\in\mathbb{R}, f:R→Rf:\mathbb{R}\to\mathbb{R} satisfy f(x+y)=f(x)+2y2+y+αxyf(x+y)=f(x)+2y^2+y+\alpha xy for all x,y∈Rx,y\in\mathbb{R}. If f(0)=−1f(0)=-1 and f(1)=2f(1)=2, then the value of ∑n=15(α+f(n))\displaystyle\sum_{n=1}^{5}\big(\alpha+f(n)\big) is
A110110
B140140correct
C150150
D170170
Solution
Step 1: Put x=0x=0 in the relation: f(y)=f(0)+2y2+y=−1+2y2+yf(y)=f(0)+2y^2+y=-1+2y^2+y, so f(x)=2x2+x−1f(x)=2x^2+x-1. Step 2: Determine α\alpha using the fact that the relation holds for all x,yx,y. Expanding both sides with f(x)=2x2+x−1f(x)=2x^2+x-1:
f(x+y)=2(x+y)2+(x+y)−1=2x2+4xy+2y2+x+y−1,f(x+y)=2(x+y)^2+(x+y)-1=2x^2+4xy+2y^2+x+y-1,
f(x)+2y2+y+αxy=2x2+x−1+2y2+y+αxy.f(x)+2y^2+y+\alpha xy=2x^2+x-1+2y^2+y+\alpha xy.
 Comparing the xyxy terms: 4xy=αxy⇒α=44xy=\alpha xy\Rightarrow\alpha=4. (Check: f(1)=2+1−1=2f(1)=2+1-1=2 ✓.) Step 3: With α=4\alpha=4 and f(n)=2n2+n−1f(n)=2n^2+n-1, α+f(n)=4+2n2+n−1=2n2+n+3\alpha+f(n)=4+2n^2+n-1=2n^2+n+3. Step 4: ∑n=15(2n2+n+3)=2∑n=15n2+∑n=15n+∑n=153=2(55)+15+15=140.\displaystyle\sum_{n=1}^{5}(2n^2+n+3)=2\sum_{n=1}^{5}n^2+\sum_{n=1}^{5}n+\sum_{n=1}^{5}3=2(55)+15+15=140. Correct answer: (2)
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