Question

(50 PONTOS) Dada uma sequência numérica $a_{n},$ definamos o operador primeira diferença posterior:

$\Delta(a_{n})=a_{n+1}-a_{n},\;\;\;\;n\in\mathbb{N}$


Dadas duas sequências numéricas $a_{n}$ e $b_{n},$ e $k$ uma constante real, mostre que valem as seguintes propriedades operatórias:

$(1)\;\;\Delta(k\cdot a_{n})=k\cdot \Delta(a_{n})\\ \\ \\ (2)\;\;\Delta(a_{n}+b_{n})=\Delta(a_{n})+\Delta(b_{n})\\ \\ \\ (3)\;\;\Delta(a_{n}\cdot b_{n})=\Delta(a_{n})\cdot b_{n}+a_{n+1}\cdot \Delta(b_{n})\\ \\ \text{ou equivalentemente,\;\;}\Delta(a_{n}\cdot b_{n})=\Delta(a_{n})\cdot b_{n+1}+a_{n}\cdot \Delta(b_{n})\\ \\ \\ (4)\;\;\Delta\left(\dfrac{a_{n}}{b_{n}} \right )=\dfrac{\Delta(a_{n})\cdot b_{n}-a_{n}\cdot \Delta(b_{n})}{b_{n+1}\cdot b_{n}},\;\;\;\;\text{ se }b_{n}\neq 0, \forall n\in \mathbb{N}$

Answer 1

(1)

Definindo $b_{n}=k\cdot a_{n}$, temos que

$\Delta(k\cdot a_{n})=\Delta(b_{n})=b_{n+1}-b_{n}\\\\\Delta(k\cdot a_{n})=k\cdot a_{n+1}-k\cdot a_{n}\\\\\Delta(k\cdot a_{n})=k\cdot(a_{n+1}-a_{n})\\\\\boxed{\boxed{\Delta(k\cdot a_{n})=k\cdot\Delta(a_{n})}}$

(2)

Definindo $c_{n}=a_{n}+b_{n}$, temos, por definição,

$\Delta(a_{n}+b_{n})=\Delta(c_{n})=c_{n+1}-c_{n}\\\\\Delta(a_{n}+b_{n})=(a_{n+1}+b_{n+1})-(a_{n}+b_{n})\\\\\Delta(a_{n}+b_{n})=(a_{n+1}-a_{n})+(b_{n+1}-b_{n})\\\\\boxed{\boxed{\Delta(a_{n}+b_{n})=\Delta(a_{n})+\Delta(b_{n})}}$

(3)

Definindo $c_{n}=a_{n}\cdot b_{n}$, temos que

$\Delta(a_{n}\cdot b_{n})=\Delta(c_{n})=c_{n+1}-c_{n}\\\\\Delta(a_{n}\cdot b_{n})=a_{n+1}\cdot b_{n+1}-a_{n}\cdot b_{n}\\\\\Delta(a_{n}\cdot b_{n})=(a_{n+1}-a_{n}+a_{n})\cdot b_{n+1}-a_{n}\cdot b_{n}\\\\\Delta(a_{n}\cdot b_{n})=(a_{n+1}-a_{n})\cdot b_{n+1}+a_{n}\cdot b_{n+1}-a_{n}\cdot b_{n}\\\\\Delta(a_{n}\cdot b_{n})=\Delta(a_{n})\cdot b_{n+1}+a_{n}\cdot(b_{n+1}-b_{n})\\\\\boxed{\boxed{\Delta(a_{n}\cdot b_{n})=\Delta(a_{n})\cdot b_{n+1}+a_{n}\cdot\Delta(b_{n})}}$

(4)

Definindo $c_{n}=\dfrac{a_{n}}{b_{n}}$, temos então

$\Delta\left(\dfrac{a_{n}}{b_{n}}\right)=\Delta(c_{n})=c_{n+1}-c_{n}\\\\\\\Delta\left(\dfrac{a_{n}}{b_{n}}\right)=\dfrac{a_{n+1}}{b_{n+1}}-\dfrac{a_{n}}{b_{n}}\\\\\\\Delta\left(\dfrac{a_{n}}{b_{n}}\right)=\dfrac{a_{n+1}\cdot b_{n}}{b_{n+1}\cdot b_{n}}-\dfrac{a_{n}\cdot b_{n+1}}{b_{n+1}\cdot b_{n}}\\\\\\\Delta\left(\dfrac{a_{n}}{b_{n}}\right)=\dfrac{a_{n+1}\cdot b_{n}-a_{n}\cdot b_{n+1}}{b_{n+1}\cdot b_{n}}$


$\Delta\left(\dfrac{a_{n}}{b_{n}}\right)=\dfrac{(a_{n+1}-a_{n}+a_{n})\cdot b_{n}-a_{n}\cdot b_{n+1}}{b_{n+1}\cdot b_{n}}\\\\\\\Delta\left(\dfrac{a_{n}}{b_{n}}\right)=\dfrac{(a_{n+1}-a_{n})\cdot b_{n}+a_{n}\cdot b_{n}-a_{n}\cdot b_{n+1}}{b_{n+1}\cdot b_{n}}\\\\\\\Delta\left(\dfrac{a_{n}}{b_{n}}\right)=\dfrac{\Delta(a_{n})\cdot b_{n}+a_{n}\cdot(b_{n}-b_{n+1})}{b_{n+1}\cdot b_{n}}\\\\\\\Delta\left(\dfrac{a_{n}}{b_{n}}\right)=\dfrac{\Delta(a_{n})\cdot b_{n}-a_{n}\cdot(b_{n+1}-b_{n})}{b_{n+1}\cdot b_{n}}$

$\boxed{\boxed{\Delta\left(\dfrac{a_{n}}{b_{n}}\right)=\dfrac{\Delta(a_{n})\cdot b_{n}-a_{n}\cdot\Delta(b_{n})}{b_{n+1}\cdot b_{n}}}}$