Q41 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics
$f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x+y)=f(x)+12 y, \forall x, y \in \mathbb{R}$ If $f(1)=6$, then $\sum_{r=1}^n f(r)=$
- A$n^2$
- B$5 n^2$
- Correct answerC$6 n^2$
- D$\frac{3 n(n+1)}{2}$
Solution
$f^{\prime}(x)=12$, Given $f(1)=6$
$f(2)=f(1)+12(1)=6+12=18$
Similarly, $f(3)=30 ; f(4)=42$ $\begin{aligned} & \sum_{r=1}^n f(r)=6+18+30+42 \ldots n \text { terms } \\ & =6(1+3+5+7 \ldots . n \text { terms })=6 . n^2 \end{aligned}$
Similarly, $f(3)=30 ; f(4)=42$ $\begin{aligned} & \sum_{r=1}^n f(r)=6+18+30+42 \ldots n \text { terms } \\ & =6(1+3+5+7 \ldots . n \text { terms })=6 . n^2 \end{aligned}$
Hence, the correct answer is option: 'C'.