Question

If $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ satisfies $f(x+y)=f\left(x\right)+f\left(y\right)$, for all $x,y\in R$, $\mathrm{f}\left(1\right)=\frac{3}{2}$, then $\underset{\mathrm{r}=1}{\overset{\mathrm{n}}{\sum \mathrm{f}\left(\mathrm{r}\right)}}$is

• A. $\left(\frac{3}{2}\right)\mathrm{n}(\mathrm{n}+1)$
• B. $\left[\frac{3\left(\mathrm{n}+1\right)}{2}\right]$
• C. $\left[\frac{3\mathrm{n}\left(\mathrm{n}+1\right)}{4}\right]$
• D. $\frac{3\mathrm{n}}{4}$

Answer 1

The correct option is C

$\left[\frac{3\mathrm{n}\left(\mathrm{n}+1\right)}{4}\right]$

Finding $\underset{\mathrm{r}=1}{\overset{\mathrm{n}}{\sum \mathrm{f}\left(\mathrm{r}\right)}}$:

We have: $f\left(1\right)=\frac{3}{2}$ , $f(x+y)=f\left(x\right)+f\left(y\right)$

$\because \mathrm{f}\left(1\right)=\frac{3}{2}$

$\begin{array}{rcl}\therefore \mathrm{f}\left(2\right)& =& \mathrm{f}(1+1)\\ & =& \mathrm{f}\left(1\right)+\mathrm{f}\left(1\right)\\ & =& \frac{3}{2}+\frac{3}{2}\\ & =& \frac{6}{2}\end{array}$

Similarly,

$\begin{array}{rcl}\therefore \mathrm{f}\left(3\right)& =& \mathrm{f}(2+1)\\ & =& \mathrm{f}\left(2\right)+\mathrm{f}\left(1\right)\\ & =& \frac{6}{2}+\frac{3}{2}\\ & =& \frac{9}{2}\end{array}$

$\begin{array}{rcl}\therefore \mathrm{f}\left(4\right)& =& \mathrm{f}(1+3)\\ & =& \mathrm{f}\left(1\right)+\mathrm{f}\left(3\right)\\ & =& \frac{3}{2}+\frac{9}{2}\\ & =& \frac{12}{2}\end{array}$ and so on.

$\therefore \mathrm{f}\left(\mathrm{n}\right)=\frac{3}{2}+\frac{\left((\mathrm{n}-1)3\right)}{2}$

Now, $\underset{\mathrm{r}=1}{\overset{\mathrm{n}}{\sum \mathrm{f}\left(\mathrm{r}\right)}}=\mathrm{f}\left(1\right)+\mathrm{f}\left(2\right)+\mathrm{f}\left(3\right)+....\mathrm{f}\left(\mathrm{n}\right)$

$=\frac{3}{2}+\frac{6}{2}+\frac{9}{2}+\frac{12}{2}+....+\frac{(\mathrm{n}-1)3}{2}$

$$\begin{gathered} =\frac{3}{2}(1+2+3+4+.....\mathrm{n}) \\ =\frac{3\mathrm{n}(\mathrm{n}+1)}{4} \end{gathered}$$

Hence, the correct option is an option (c).