已知数列an的前n项和Sn=n^2+2n，求通项an 求和1/a1a2+1/a2a3+1/a3a4…1/anan+1

显示全部楼层 · 2012-7-9 20:40:43

an=S(n)-S(n-1)=n^2+2n-[(n-1)^2+2(n-1)]=n^2+2n-(n^2-2n+1+2n-2)=n^2+2n-n^2+1=2n+11/a1a2+1/a2a3+1/a3a4…1/anan+1=1/(3*5)+1/(5*7)+1/(7*9)+...+1/anan+1=1/2[(1/3-1/5)+(1/5-1/7)+(1/7-1/9)+...+(1/(2n+1)-1/(2n+3))]=1/2([1/3-1/(2n+3)]=1/2*[(2n+3-3)/3(2n+3)]=n/3(2n+3)...

千问 · 2012-7-9 20:40:43

Sn=n^2+2nS(n-1)=(n-1)^2+2(n-1)=n^2-1an=2n+11/a1a2+1/a2a3+1/a3a4…1/anan+1=1/(3*5)+1/(5*7)+1/(7*9)+...+1/anan+1=1/2[(1/3-1/5)+(1/5-1/7)+(1/7-1/9)+...+(1/(2n+1)-1/(2n+3))]...