Q43 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

If $\mathrm{f}(\mathrm{x})$ is a function such that $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})$ and $\mathrm{f}(1)$ $=7$ then $\sum_{r=1}^n \mathrm{f}(\mathrm{r})=$

  • A
    $\frac{7 n}{2}$
  • B
    $\frac{7(n+1)}{2}$
  • C
    $7 n(n+1)$
  • Correct answer
    D
    $\frac{7 \mathrm{n}(\mathrm{n}+1)}{2}$

Solution

$f(x+y)=f(x)+f(y)$ $\begin{aligned} & f(1)=7 \\ & \text { put } x=1, \mathrm{y}=0 \\ & f(1)=f(1)+f(0) \Rightarrow f(0)=0 \\ & \text { Put } x=1, y=1 \\ & f(2)=2 f(1)=2.7=14 \\ & \text { and } f(3)=f(2)+f(1)=14+7=21 \\ & \therefore \sum_{r=1}^n f(r)=f(1)+f(2)+\ldots . .+f(n) \\ & =7+14+21+\ldots . n \text { terms } \\ & =7(1+2+3+\ldots . n \text { terms }) \\ & =7 . \frac{n(n+1)}{2}\end{aligned}$

Hence, the correct answer is option: 'D'.

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