Question
Let for some $\alpha \in \mathbb{R},f:\mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying $f(x + y) = f(x) + 2y^{2} + y + \alpha xy$ for all $x,y \in \mathbb{R}$. If $f(0) = - 1$ and $f(1\,) = 2$, then the value of $\sum_{n = 1}^{5}\mspace{2mu}(\alpha + f(n))$ is :
Options
Solution
Sol. [Key Concept: Put $x = 0$ to find $f(y)$; put $x = y = 1$ to get $\alpha$, then sum the closed form $\sum_{n = 1}^{5}(\alpha + f(n))$.]
$$f(x + y) = f(x) + 2y^{2} + y + \alpha xy$$
Put $x = 0$ in eq.(1)
$$\Rightarrow f(y) = - 1 + 2y^{2} + y $$Now put $x = y = 1$ in eq. (1)
$${\Rightarrow f(2) = f(1) + 3 + \alpha }{\Rightarrow 9 = 2 + 3 + \alpha \Rightarrow \alpha = 4 }$$Now $\sum_{n = 1}^{5}\mspace{2mu}(\alpha + f(n)) = \sum_{n = 1}^{5}\mspace{2mu}\left( 2y^{2} + y + 3 \right)$
$$= \frac{2 \times 5 \times 6 \times 11}{6} + \frac{5 \times 6}{2} + 3 \times 5 = 140 $$
Create a free account to view solution
View Solution Free