高二数学达人进！数列{an}的通项公式an=n(n+1)/2,求数列{an}的前n项和Sn。回答得好重重有赏！

显示全部楼层 · 2011-10-6 21:58:30

∵an=n(n+1)/2=n^2/2+n/2∴Sn=a1+a2+...+an
=(1^2/2 +1/2)+(2^2/2+2/2)+...+(n^2/2+n/2)
=1/2(1+2^2+...+n^2)+1/2(1+2+...+n)
=1/2*1/6n(n+1)(2n+1)+1/2*n(n+1)/2
=n(n+1)/12*[(2n+1）+3]
=n(n+1)(n+2)/6...

千问 · 2011-10-6 21:58:30

解 Sn=a1+a2+..+an
=1*2/2+2*2/2+....+n(n+1)/2
=(1+2+...+n)/2+(1^2+2^2+...+n^2)/2
=n（n+1)/4 +1/2n(n+1)(2n+1)/6
=n(n+1)(2n+1+3)/12
=n(n+1)(n+...

千问 · 2011-10-6 21:58:30

an=n(n+1)/2=n2/2+n/2所以Sn=1/2【(12+22+。。。。。。n2）+（1+2+。。。。。n）】
=1/2【n（n+1）（2n+1）/6+(n+1)n/2】
=1/2【n（n+1）（2n+4）/6】
=n（n+1）（n+2）/6...