Question

Mulțumesc anticipat!

Answer 1

$\it f(x)=x-a\\ \\ a)\ P(3a-4,\ 0)\in Gf \Rightarrow f(3a-4)=0 \Rightarrow 3a-4-a=0\Rightarrow2a=4\Rightarrow a=2\\ \\ b)\ a=2\Rightarrow f(x)=x-2\\ \\ f(1)+f(2)+f(3)+f(4)+\ ...\ +f(n)=-1+0+1+2+\ ...\ +n-2=\\ \\ =-1+\dfrac{(n-2)(n-1)}{2}=\dfrac{n^2-3n+2}{2}-1=\dfrac{n^2-3n+2-2}{2}=\dfrac{n(n-3)}2{}$

Answer 2

$f:\mathbb{R}\to \mathbb{R},\quad f(x) = x-a\\ \\ a)\quad P(3a-4,\,0)\in Gf\,\Rightarrow\, f(3a-4) = 0\, \Leftrightarrow \,\\ \Leftrightarrow \, (3a-4)-a = 0\,\Leftrightarrow \, 3a-a-4 = 0\, \Leftrightarrow\\\Leftrightarrow\, 2a = 4 \, \Leftrightarrow\,\boxed{a = 2}$

$\displaystyle\left.\begin{aligned}b)\quad f(1)+f(2)+...+f(n) &=\sum\limits_{k=1}^{n}f(k)\\&=\sum\limits_{k=1}^n(k-2)\\ &= \sum\limits_{k=1}^n k - \sum\limits_{k=1}^n 2\\ &= \dfrac{n(n+1)}{2}-2n\\ &= \dfrac{n(n+1)-4n}{2}\\ &= \dfrac{n^2+n-4n}{2}\\&= \boxed{\dfrac{n(n-3)}{2}}\displaystyle \end{aligned}\right.$