Question

已知数列{an}的通项公式为an=一分之(2n+1)(2n+3),求数列的前N项和

Answer 1

解:an=1/(2n十1)(2n十3):Sn二a1十α2十α3十…十αn-1十αn二1/2(1/3-1/5十1/5-1/7十…十1/2n+1-1/2n十3)=1/2(1/3-1/2n十3)二1/2ⅹ2n/6n十9=n/6n十9

Answer 2

an
= 1/[(2n+1)(2n+3)]
=(1/2)[ 1/(2n+1)-1/(2n+3)]
Sn
=a1+a2+...+an
=(1/2)[ 1/3 -1/(2n+3)]