Question

{n(n+1)(n+2)}分之一怎样求和

Answer 1

1/(n(n+1)(n+2)

=(1/2)*2/n(n+1)(n+2)

=(1/2)*[(n+2)-n]/n(n+1)(n+2)

=(1/2)*[(n+2)/n(n+1)(n+2)-n/n(n+1)(n+2)]

=(1/2)*[1/n(n+1)-1/(n+1)(n+2)]

所以1/1*2*3+……+1/n(n+1)(n+2)

=(1/2)*[1/1*2-1/2*3+……+1/n(n+1)-1/(n+1)(n+2)]

=(1/2)*[1/1*2-1/(n+1)(n+2)]

=(n²+3n)/(4n²+12n+8)