¿Cómo se puede calcular $(1 + \frac{1}{2})(1 + \frac{1}{4})(1 + \frac{1}{8})(1 + \frac{1}{{16}}) \cdots (1 + \frac{1}{{{2^{10}}}})$ ?
Respuesta
¿Demasiados anuncios?$$(1 + \frac{1}{2})(1 + \frac{1}{4})(1 + \frac{1}{8})(1 + \frac{1}{{16}}) \cdots (1 + \frac{1}{{{2^{10}}}})$$ $$=(1+\frac{512}{1024})(1+\frac{256}{1024})(1+\frac{128}{1024})(1+\frac{64}{1024})(1+\frac{32}{1024})(1+\frac{16}{1024})(1+\frac{8}{1024})(1+\frac{4}{1024})(1+\frac{2}{1024})(1+\frac{1}{1024})$$ $$=(\frac{1536}{1024})(\frac{1280}{1024})(\frac{1152}{1024})(\frac{1088}{1024})(\frac{1056}{1024})(\frac{1040}{1024})(\frac{1032}{1024})(\frac{1028}{1024})(\frac{1026}{1024})(\frac{1025}{1024})$$ $$=\frac{1536*1280*1152*1088*1056*1040*1032*1028*1026*1025}{1024^{10}}$$ $$=\frac{85817142703524375}{36028797018963968}$$
;-)