题目

猜想sn=1+1/(1+2)+1/(1+2+3)+…+1/(1+2+3+…+n)的表达式,并用数学归纳法证明

提问时间：2020-08-07

答案

1/(1+2+3+…+n) =2/(n(n+1)) =2/(1/n-1/(n+1))
sn =2(1/1-1/2+1/2-1/3+.+1/n-1/(n+1))
=2(1-1/(n+1))
=2n/(n+1)
当n＝1时,
s1=2*1/(1+1) =1 成立
当n＝k时,假设成立
sk＝2k/(k+1)
当n＝k+1是
s(k+1)= sk +1/(1+2+3+…+(k+1))
=2k/(k+1)+ 2/((k+1)(k+2))
=2(k/(k+1)+1/(k+1)-1/(k+2))
=2(1-1/(k+2))
=2(k+1)/((k+1)+1) 成立