Question

simplificati fractia 1+2+...+50 supra 4+8+...+100

Answer 1

$\dfrac{1+2+...+50}{4+8+...+100} = \dfrac{1+2+...+50}{4\cdot(1+2+...+25)} = \dfrac{ \dfrac{50\cdot 51}{2} }{4\cdot \dfrac{25\cdot 26}{2} } = \dfrac{{ \dfrac{50\cdot 51}{2} }}{2\cdot 25\cdot 26} = \\ \\ = \dfrac{50\cdot 51}{2\cdot 2\cdot 25\cdot 26} = \dfrac{50\cdot 51}{2\cdot 50\cdot 26} = \dfrac{51}{2\cdot 26} = \dfrac{51}{52}$

Answer 2

_ 1+ 2+ ... + 50__  =
   4+ 8+ ... + 100


_ ___50·51:2________  =
  4( 1+ 2+ ... + 25)


 ___50·51:2___ =
    4·25·26:2 


__25·51___ =
  4·25·13


_51_
  52