[tex]\dfrac{1+4+7+10+...+301}{1+5+9+13+...+401} = \\ \\ = \dfrac{1+(5-1)+(9-2)+(13-3)+...+(401-100)}{1+5+9+13+...+401} =\\ \\ = \dfrac{1+5+9+13+...+401-(1+2+3+...+100)}{1+5+9+13+...+401} = \\ \\ =1 - \dfrac{1+2+3+...+100}{1+(2+3)+(3+4)+(5+6)+...+(200+201)} = \\ \\ = 1 - \dfrac{100\cdot 101}{201\cdot 202} =1-\dfrac{100}{201\cdot 2} = \dfrac{302}{402}=\dfrac{151}{201}[/tex]
