Let f:RR satisfy f(x+y)=f(x)+f(y), x,yR.

Question

Let $f:\mathbb{R}	o\mathbb{R}$ satisfy $f(x+y)=f(x)+f(y)$, $\forall x,y\in\mathbb{R}$. If $f(1)=2$ and $g(n)=\displaystyle\sum_{k=1}^{n-1}f(k)$, $n\in\mathbb{N}$, then the value of $n$ for which $g(n)=20$ is:

Accepted Answer

$f(k)=2k$, so $g(n) = 2+4+\cdots+2(n-1) = (n-1)n = n^2-n$.

$$n^2-n=20 \implies (n-5)(n+4)=0 \implies n=5.$$

Solutions · 1