Periodo de la división del número 1/998001
$$\begin{align}&\text{El número}\\&\\&\frac{1}{998001}=0.\,000\,001\,002\,003\,004\,005 ...996\,997\,999...\\&\\&\text{¿Por qué?}\\&\\&\text{Sabemos que la suma infinita de esta progresión geométrica es}\\&\\&1+x+x^2+x^3 +x^4+x^5+...= \frac{1}{1-x}\quad para\; |x|<1\\&\\&entonces\\&\\&1+2x+3x^2+4x^3+5x^4+...=\\&\\&1+x+x^2+x^3+x^4+x^5+...+\\&\;\;\;\;\;\; x+x^2+x^3+x^4+x^5+...+\\&\qquad\qquad\;\,+x^3+x^4+x^5+...=\\&\\&\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+ \frac{x^3}{1-x}+...=\\&\\&\frac{1}{1-x}(1+x+x^2+x^3+....)=\\&\\&\frac{1}{1-x}·\frac{1}{1-x}= \frac{1}{(1-x)^2}\\&\\&\text{Tomando x=0.001}\\&\\&1+0.001+0.000\,002+0.000\,000\,003+0.000\,000\,000\,004+...=\\&\\&1.001002003004005006...=\\&\\&\frac{1}{(1-0.001)^2}= \frac{1}{0.999^2}= \frac{1000000}{998001}\\&\\&\text{luego}\\&\\&\frac{1.001\,002\,003\,004\,005\,006...}{1000000}=\frac{1}{998001}\\&\\&0.000\,001\,002\,003\,004\,005\,006...=\frac{1}{998001}\\&\\&\text{¿Por qué desaparece el 998?}\\&\\&\text{La suma al final es:}\\&...997\,998\,999 \\&+...0\,000\,001\,000\\&+...0\,000\,001\,001\\&+...0\,000\,001\,002\\&+\\&\text{__________________}\\&...997\,999\,002\,003\\&\\&\\&\end{align}$$No es una pregunta, es solo para hacer un meme con latex.
1 Respuesta
Respuesta de tmoli
1
Perdón, tuve un fallo al final lo voy a corregir.
$$\begin{align}&\text{El número}\\&\\&\frac{1}{998001}=0.\,000\,001\,002\,003\,004\,005 ...996\,997\,999...\\&\\&\text{¿Por qué?}\\&\\&\text{Sabemos que la suma infinita de esta progresión geométrica es}\\&\\&1+x+x^2+x^3 +x^4+x^5+...= \frac{1}{1-x}\quad para\; |x|<1\\&\\&entonces\\&\\&1+2x+3x^2+4x^3+5x^4+...=\\&\\&1+x+x^2+x^3+x^4+x^5+...+\\&\;\;\;\;\;\; x+x^2+x^3+x^4+x^5+...+\\&\qquad\qquad\;\,+x^3+x^4+x^5+...=\\&\\&\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+ \frac{x^3}{1-x}+...=\\&\\&\frac{1}{1-x}(1+x+x^2+x^3+....)=\\&\\&\frac{1}{1-x}·\frac{1}{1-x}= \frac{1}{(1-x)^2}\\&\\&\text{Tomando x=0.001}\\&\\&1+0.001+0.000\,002+0.000\,000\,003+0.000\,000\,000\,004+...=\\&\\&1.001002003004005006...=\\&\\&\frac{1}{(1-0.001)^2}= \frac{1}{0.999^2}= \frac{1000000}{998001}\\&\\&\text{luego}\\&\\&\frac{1.001\,002\,003\,004\,005\,006...}{1000000}=\frac{1}{998001}\\&\\&0.000\,001\,002\,003\,004\,005\,006...=\frac{1}{998001}\\&\\&\text{¿Por qué desaparece el 998?}\\&\\&\text{La suma al final es:}\\&...997\,998\,999 \\&+...0\,000\,001\,000\\&+...0\,000\,000\,001\,001\\&+...0\,000\,000\,000\,001\,002\\&+\\&\text{________________________}\\&...997\,999\,000\,001\,002\,003...\\&\\&\\&\end{align}$$
