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