Question

5- (FRB-BA) Seja f: N R, uma função , com a seguinte propriedade.

f(1) = 3 e f(n + 1) = 2/3 f (n) – ½. Determine o valor de f(2) – f(3).​

Answer 1

Resposta:

Explicação passo-a-passo:

Temos:

$f(1)=3\\\\e\\\\f(n+1)=\dfrac{2}{3}f(n)-\dfrac{1}{2}$

Para n=1

$f(1+1)=\dfrac{2}{3}f(1)-\dfrac{1}{2}\\\\\\f(2)=\dfrac{2}{3}.3-\dfrac{1}{2}\\\\\\f(2)=2-\dfrac{1}{2}\\\\\\f(2)=\dfrac{4-1}{2}\\\\\\\boxed{f(2)=\dfrac{3}{2}}$

Para n=2

$f(2+1)=\dfrac{2}{3}f(2)-\dfrac{1}{2}\\\\\\f(3)=\dfrac{2}{3}.\dfrac{3}{2} -\dfrac{1}{2}\\\\f(3)=1 -\dfrac{1}{2}\\\\\\f(3)=\dfrac{2-1}{2}\\\\\\\boxed{f(3)=\dfrac{1}{2}}\\\\\\f(2)-f(3)=\dfrac{3}{2}-\dfrac{1}{2}=\dfrac{2}{2}\\\\\\\boxed{f(2)-f(3)=1}$