Question

Mostre que a igualdade abaixo é verdadeira

$\mathsf{\displaystyle\sum_{k=1}^\infty\dfrac{F_k}{10^{k+1}}=\dfrac{1}{89}}$

Onde Fn é a sequência de Fibonacci.

$\mathsf{F_n = \dfrac{1}{\sqrt{5}}\cdot\Big[\Big(\dfrac{1+\sqrt{5}}{2}\Big)^n - \Big(\dfrac{1-\sqrt{5}}{2}\Big)^n\Big]}}$

Sabendo que

$\mathsf{\displaystyle\sum_{k=p}^{\infty} x^k = \dfrac{x^p}{1-x}~~~se~~|x|\ \textless \ 1}$

________

Por favor responder de forma detalhada.

Answer 1

Inicialmente, temos que $F_k=\dfrac{x^{k}-y^{k}}{x-y}$, onde $x=\dfrac{1+\sqrt{5}}{2}$ e $y=\dfrac{1-\sqrt{5}}{2}$.

Note que $x+y=1$, $x-y=\sqrt{5}$ e $xy=-1$

Colocando $\dfrac{1}{10}$ em evidência:

$\displaystyle\sum_{k=1}^{\infty} \frac{F_k}{10^{k+1}}=\dfrac{1}{10}\cdot\displaystyle\sum_{k=1}^{\infty} \frac{F_k}{10^{k}}$

Substituindo $F_k$ por $\dfrac{x^{k}-y^{k}}{x-y}$:

$\displaystyle\sum_{k=1}^{\infty} \frac{F_k}{10^{k+1}}=\dfrac{1}{10}\cdot\displaystyle\sum_{k=1}^{\infty} \frac{\frac{x^{k}-y^{k}}{x-y}}{10^k}}=\dfrac{1}{10}\cdot\displaystyle\sum_{k=1}^{\infty} \frac{x^{k}-y^{k}}{10^{k}(x-y)}}$

Colocando $\dfrac{1}{x-y}$ em evidência:

$\displaystyle\sum_{k=1}^{\infty} \frac{F_k}{10^{k+1}}=\dfrac{1}{10(x-y)}\cdot\displaystyle\sum_{k=1}^{\infty} \frac{x^{k}-y^{k}}{10^{k}}}$

Mas, $\dfrac{x^{k}-y^{k}}{10^{k}}=\dfrac{x^{k}}{10^{k}}-\dfrac{y^{k}}{10^{k}}=\left(\dfrac{x}{10}\right)^{k}-\left(\dfrac{y}{10}\right)^{k}$, então

$\displaystyle\sum_{k=1}^{\infty} \frac{F_k}{10^{k+1}}=\dfrac{1}{10(x-y)}\cdot\left[\displaystyle\sum_{k=1}^{\infty} \left(\frac{x}{10}\right)^{k}-\left(\frac{y}{10}\right)^{k}\right]$

Como $\left|\dfrac{x}{10}\right|<0$ e $\left|\dfrac{y}{10}\right|<0$, pois $-1<\dfrac{x}{10}=\dfrac{1+\sqrt{5}}{20}<1$ e $-1<\dfrac{y}{10}=\dfrac{1-\sqrt{5}}{20}<1$, temos que:

$\displaystyle\sum_{k=1}^{\infty} \left(\frac{x}{10}\right)^{k}-\left(\frac{y}{10}\right)^{k}\right=\displaystyle\sum_{k=1}^{\infty} \left(\frac{x}{10}\right)^{k}-\displaystyle\sum_{k=1}^{\infty}\left(\frac{y}{10}\right)^{k}\right$

Pelo enunciado, $\displaystyle\sum_{k=p}^{\infty} x^{k}=\dfrac{x^{p}}{1-x}$, assim:

$\displaystyle\sum_{k=1}^{\infty} \left(\frac{x}{10}\right)^{k}=\dfrac{\frac{x}{10}}{1-\frac{x}{10}}=\dfrac{\frac{x}{10}}{\frac{10-x}{10}}=\dfrac{x}{10-x}$

$\displaystyle\sum_{k=1}^{\infty} \left(\frac{y}{10}\right)^{k}=\dfrac{\frac{y}{10}}{1-\frac{y}{10}}=\dfrac{\frac{y}{10}}{\frac{10-y}{10}}=\dfrac{y}{10-y}$

Desse modo, $\displaystyle\sum_{k=1}^{\infty} \left(\frac{x}{10}\right)^{k}-\displaystyle\sum_{k=1}^{\infty}\left(\frac{y}{10}\right)^{k}\right=\dfrac{x}{10-x}-\dfrac{y}{10-y}$

Logo:

$\displaystyle\sum_{k=1}^{\infty} \frac{F_k}{10^{k+1}}=\dfrac{1}{10(x-y)}\cdot\left[\displaystyle\sum_{k=1}^{\infty} \left(\frac{x}{10}\right)^{k}-\left(\frac{y}{10}\right)^{k}\right]$

$\displaystyle\sum_{k=1}^{\infty} \frac{F_k}{10^{k+1}}=\dfrac{1}{10(x-y)}\cdot\left(\dfrac{x}{10-x}-\dfrac{y}{10-y}\right)$

$\displaystyle\sum_{k=1}^{\infty} \frac{F_k}{10^{k+1}}=\dfrac{1}{10(x-y)}\cdot\dfrac{10x-xy-10y+xy}{100-10x-10y+xy}$

$\displaystyle\sum_{k=1}^{\infty} \frac{F_k}{10^{k+1}}=\dfrac{1}{10(x-y)}\cdot\dfrac{10(x-y)}{100-10(x+y)+xy}$

Lembrando que $x+y=1$ e $xy=-1$ obtemos:

$\displaystyle\sum_{k=1}^{\infty} \frac{F_k}{10^{k+1}}=\dfrac{10(x-y)}{10(x-y)\cdot(100-10-1)}$

$\boxed{\displaystyle\sum_{k=1}^{\infty} \frac{F_k}{10^{k+1}}=\dfrac{1}{89}}$