１，１／１＋２，１／１＋２＋３，１／１＋２＋３＋４....的前n 項的Sn

熱心網友

因：1+2 = 2(1+2)/2 = 2×3/21+2+3 = 3(1+3)/2 =3×4/21+2+3+4 = 4(1+4)/2 =4×5/21+2+3+……+n = n(1+n)/2所以原數列前n項的和Sn=1+2/(2×3)+2/(3×4)+ … + 2/[n×(n+1)]。=2[(1-1/2)+(1/2-1/3)+(1/3-1/4+(1/4-1/5)+...+1/n-1/(n+1)]=2[1–1/(n+1)]=2n/(n+1)

熱心網友

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