Question

Halle el valor de A
A=[(1!+2!+3!)(2!+3!+4!)(3!+4!+5!)......(18!+19!+20!)]/[20!.19!.18!.....2!.1!]

Answer 1

$\dfrac{(1!+2!+3!)(2!+3!+4!)(3!+4!+5!)......(18!+19!+20!)}{20!.19!.18!.....2!.1!}\\\\\dfrac{1!(1+2+2.3)2!(1+3+3.4)3!(1+4+4.5)...18!(1+19+19.20)}{20!.19!.18!.....2!.1!}\\\\\dfrac{1!.2!.3!.4!....18!(9)(16)(25)...(400)}{20!.19!.18!.....2!.1!}\\\\\dfrac{(9)(16)(25)...(400)}{20!.19!}\\\\\dfrac{3^{2}4^{2}5^{2}.....20^{2} }{20!.19!}.\dfrac{1^{2}2^{2}}{1^{2}2^{2}}\\\\\dfrac{1^{2}2^{2}3^{2}4^{2}5^{2}.....20^{2}}{1^{2}2^{2}.20!.19!}\\\\\dfrac{(1.2.3.4.5.....20)^{2}}{1^{2}2^{2}.20!.19!}$

$\dfrac{20!^{2}}{1^{2}2^{2}.20!.19!}\\\\\\\dfrac{20!}{4.19!}\\\\\dfrac{20.19!}{19!.4}\\\\5$

RESPUESTA

A=5