Question:
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function which satisfies $f(x+y)=f(x)+f(y), \forall x, y \in \mathbf{R}$. If $f(1)=2$ and
$g(n)=\sum_{k=1}^{(n-1)} f(k), n \in \mathbf{N}$, then the value of $n$, for which
$g(n)=20$, is :
Correct Option: 1
Solution:
Given: $f(x+y)=f(x)+f(y), \forall x, y \in R, f(1)=2$
$\Rightarrow f(2)=f(1)+f(1)=2+2=4$
$f(3)=f(1)+f(2)=2+4=6$
$f(n-1)=2(n-1)$
Now, $g(n)=\sum_{k=1}^{n-1} f(k)$
$=f(1)+f(2)+f(3)+\ldots . f(n-1)$
$=2+4+6+\ldots . .+2(n-1)=2[1+2+3+\ldots . .+(n-1)]$
$=2 \times \frac{(n-1)(n)}{2}=n^{2}-n$
$\because g(n)=20$ (given)
So, $n^{2}-n=20$
$\Rightarrow n^{2}-n-20=0 \Rightarrow(n-5)(n+4)=0$
$\Rightarrow n=5$ or $n=-4$ (not possible)