Question

3/(1-а^3)+3/(1+а^3)+6/(1+а^6)+12/(1+а^12)+24/(1+а^24)+48/(1+а^48)=?

Answer 1

$displaystylefrac{3}{1-a^3}+frac{3}{1+a^3} +frac{6}{1+a^6}+frac{12}{1+a^{12}}+frac{24}{1+a^{24}}+frac{48}{1+a^{48}}=\ \ \ =frac{6}{1-a^6} +frac{6}{1+a^6}+frac{12}{1+a^{12}}+frac{24}{1+a^{24}}+frac{48}{1+a^{48}}=\ \ \ =frac{12}{1-a^{12}}+ frac{12}{1+a^{12}}+frac{24}{1+a^{24}}+frac{48}{1+a^{48}}=\ \ =frac{24}{1-a^{24}}+ frac{24}{1+a^{24}}+frac{48}{1+a^{48}}=frac{48}{1-a^{48}} +frac{48}{1+a^{48}}=frac{96}{1-a^{96}}$