Question

F(x)=ax+b
a) Determinati a,b astfel incat M(0,-1) si N(-1,3) sa apartina graficului f
b)Pentru a=2 si b= -1 calculati f(1/100)*f(1/99)*f(1/98)*...*f(1/3)*f(1/2)
c) a=2 si b= -1 calculati f(1/1*2)+f(1/2*3)+f(1/3*4)+...+f(1/99*100)

Answer 1

La c iti voi posta in comentarii, am gresit undeva si inca nu am gasit greseala.

Answer 2

Am sa il fac eu pe c:

[tex]c)f\left(\dfrac{1}{1\cdot 2}\right)+f\left(\dfrac{1}{2\cdot 3}\right)+f\left(\dfrac{1}{3\cdot 4}\right)+_{\dots}+f\left(\dfrac{1}{99\cdot 100}\right)=\\ 2\cdot \left(\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+_{\dots}+\dfrac{1}{99\cdot 100}\right) \underbrace{-1-1-{\dots}-1}=\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~de\ 99\ ori\\ 2\cdot \left(\dfrac{2-1}{1\cdot 2}+\dfrac{3-2}{2\cdot 3}+\dfrac{4-3}{3\cdot 4}+_{\dots}+\dfrac{100-99}{99\cdot 100}\right)-99=\\ [/tex]
[tex]2\cdot \left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+_{\dots}+\dfrac{1}{99}-\dfrac{1}{100}\right)-99=\\ 2\cdot \left(1-\dfrac{1}{100}\right)-99=2\cdot \dfrac{99}{100}-99=\dfrac{99 -50\cdot 99}{50}=\dfrac{-49 \cdot 99}{50}[/tex]