an=n*(n+1)*(n+2)分之一 求sn=a1+a2+a3+a4+a5+…………+an为多少?求详解.

a‹n›=1/[n(n+1)(n+2)]
=(1/2)[1/n(n+1)-1/(n+1)(n+2)]
=(1/2){[1/n-1/(n+1)]-[1/(n+1)-1/(n+2)]}
故S‹n›=(1/2){[(1-1/2)-(1/2-1/3)]+[(1/2-1/3)-(1/3-1/4)]+[(1/3-1/4)-(1/4-1/5)]+[(1/4-1/5)-(1/5-1/6)]
+.+[(1/n-1/(n+1))-(1/(n+1)-1/(n+2))]}
=(1/2){(1/2-1/6)+(1/6-1/12)+(1/12-1/20)+(1/20-1/30)+.+[1/n(n+1)-1/(n+1)(n+2)]}
=(1/2)[1/2-1/(n+1)(n+2)]

=(1*1/2*1/3)+(1/2*1/3*1/4)+(1/3*1/4*1/5)+...+(1/n-2 * 1/n-1 *1/n)

an=n*(n+1)*(n+2)分之一=1/2*[1/n(n+1)-1/(n+1)(n+2)]
sn=a1+a2+a3+a4+a5+…………+an=1/2*[1/1*2-1/(n+1)(n+2)]