JEE Main 2026 — 4 April, Evening Shift — Question 40

$f(x+y)=f(x)+2 y^{2}+y+\alpha x y$

Put $\mathrm{x}=0$ in eq.(1) $\Rightarrow \mathrm{f}(\mathrm{y})=-1+2 \mathrm{y}^{2}+\mathrm{y}$ Now put $\mathrm{x}=\mathrm{y}=1$ in eq. (1) $\Rightarrow \mathrm{f}(2)=\mathrm{f}(1)+3+\alpha$ $\Rightarrow 9=2+3+\alpha \Rightarrow \alpha=4$ Now $\sum_{n=1}^{5}(\alpha+f(n))=\sum_{n=1}^{5}\left(2 y^{2}+y+3\right)$ $=\frac{2 \times 5 \times 6 \times 11}{6}+\frac{5 \times 6}{2}+3 \times 5=140$