Exercițiul 222

E.222. Să se demonstreze formula: a,(bc)‾=abc‾−a99.\overline{a,(bc)} = \dfrac{\overline{abc}-a}{99}.

MM, 24.04.2024
Soluție
Soluție:

Metoda 1: Notăm a,(bc)‾=x∣×100\overline{a,(bc)} = x \quad | \times100
abc,(bc)‾=10x∣−x\overline{abc,(bc)} = 10x \quad | -x
abc,(bc)‾−a,(bc)‾=99x\overline{abc,(bc)} - \overline{a,(bc)} = 99x
abc‾−a=99x⇒x=abc‾−a99.\overline{abc}-a=99x \Rightarrow \boxed{x=\dfrac{\overline{abc}-a}{99}}.

Metoda 2: a,(bc)‾=a+(0,bc+0,00bc+0,0000bc+… )=\overline{a,(bc)} = a + (0,bc + 0,00bc + 0,0000bc + \dots )=
=a+(bc‾102+bc‾104+bc‾106+…)== a+ \Big(\dfrac{\overline{bc}}{10^2} + \dfrac{\overline{bc}}{10^4} + \dfrac{\overline{bc}}{10^6} + \dots \Big)=
=a+bc‾102⋅lim⁡n→∞∑k=0n1102k=a+bc‾102⋅lim⁡n→∞1−1102n+21−1102==a + \dfrac{\overline{bc}}{10^2} \cdot \lim\limits_{n\to\infty} \sum\limits_{k=0}^{n} \dfrac{1}{10^{2k}}=a + \dfrac{\overline{bc}}{10^2} \cdot \lim\limits_{n\to\infty} \dfrac{1-\dfrac{1}{10^{2n+2}}}{1-\dfrac{1}{10^2}}=
=a+bc‾102⋅11−1102=a⋅102+bc‾−a99=abc‾−a99.=a + \dfrac{\overline{bc}}{10^2} \cdot \dfrac{1}{1-\dfrac{1}{10^2}}=\dfrac{a\cdot 10^2 + \overline{bc}-a}{99} = \dfrac{\overline{abc}-a}{99}.