Question Number 10216 by konen last updated on 30/Jan/17

(((√2) +(√4)+...+(√(28)))/((√3)+(√6)+...+(√(42))))=(√x) ⇒x=?

Answered by nume1114 last updated on 30/Jan/17

LHS=((Σ_(n=1) ^(14) (√(2n)))/(Σ_(n=1) ^(14) (√(3n)))) =((√2)/(√3))∙((Σ_(n=1) ^(14) (√n))/(Σ_(n=1) ^(14) (√n))) =(√(2/3)) (√(2/3))=(√x)⇒x=(2/3)

Answered by arge last updated on 04/Feb/17

para el numerador, a_1 =(√2) a_n =(√(28)) a_n =(√(2n)) n=14 para el denominador, a_1 =(√3) a_n =(√(42)) a_n =(√(2n)) n=21 (S_n )=((n(a_1 −a_n ))/2) (√x) =((7((√( 2 )) −(√(28))))/(((21)/2)((√3) −(√( 42)) ))) x= 0.74