FunctionsmediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

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

\alpha\in\mathbb{R}

f:\mathbb{R}\to\mathbb{R}

satisfy

f(x+y)=f(x)+2y^2+y+\alpha xy

for all

x,y\in\mathbb{R}

. If

f(0)=-1

and

f(1)=2

, then the value of

\displaystyle\sum_{n=1}^{5}\big(\alpha+f(n)\big)

110

140

correct

150

170

Solution

Step 1: Put

x=0

in the relation:

f(y)=f(0)+2y^2+y=-1+2y^2+y

, so

f(x)=2x^2+x-1

. Step 2: Determine

\alpha

using the fact that the relation holds for all

x,y

. Expanding both sides with

f(x)=2x^2+x-1

f(x+y)=2(x+y)^2+(x+y)-1=2x^2+4xy+2y^2+x+y-1,

f(x)+2y^2+y+\alpha xy=2x^2+x-1+2y^2+y+\alpha xy.

Comparing the

xy

terms:

4xy=\alpha xy\Rightarrow\alpha=4

. (Check:

f(1)=2+1-1=2

✓.) Step 3: With

\alpha=4

and

f(n)=2n^2+n-1

\alpha+f(n)=4+2n^2+n-1=2n^2+n+3

. Step 4:

\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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