$f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x+y)=f(x)+12 y, \forall x, y \in \mathbb{R}$ If…

$f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x+y)=f(x)+12 y, \forall x, y \in \mathbb{R}$ If $f(1)=6$, then $\sum_{r=1}^n f(r)=$
  1. $n^2$
  2. $5 n^2$
  3. $6 n^2$
  4. $\frac{3 n(n+1)}{2}$

Solution

$f^{\prime}(x)=12$, Given $f(1)=6$ $f(2)=f(1)+12(1)=6+12=18$
Similarly, $f(3)=30 ; f(4)=42$ $\begin{aligned} & \sum_{r=1}^n f(r)=6+18+30+42 \ldots n \text { terms } \\ & =6(1+3+5+7 \ldots . n \text { terms })=6 . n^2 \end{aligned}$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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