$f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x+y)=f(x)+12 y, \forall x, y \in \mathbb{R}$ If…
- $n^2$
- $5 n^2$
- $6 n^2$
- $\frac{3 n(n+1)}{2}$
Solution
Similarly, $f(3)=30 ; f(4)=42$ $\begin{aligned} & \sum_{r=1}^n f(r)=6+18+30+42 \ldots n \text { terms } \\ & =6(1+3+5+7 \ldots . n \text { terms })=6 . n^2 \end{aligned}$
Asked in: AP EAMCET 2024 (20 May Shift 2)