一个数列Sn＝n²＋2n,求1／Sn的前n项和_数学解答

一个数列Sn＝n²＋2n,求1／Sn的前n项和

人气：484 ℃　时间：2020-06-19 05:08:15

解答

∵1/Sn=1/(n²+2n)=1/n(n+2)=1/2×(1/n-1/n+2)
∴1/s1+1/s2+1/s3+.＋1/sn
=1/2(1-1/3)+1/2(1/2-1/4)+1/2(1/3-1/5)+.+1/2×(1/n-1/n+2)
=1/2(1-1/3+1/2-1/4+1/3-1/5+.+1/n-1/n+2)
=1/2[1-1/2-1/(n+1)-1/(n+2)]
=(n²-n-4)/2(n+1)(n+2)