sn=1+3+6+10+15+.共有n项,求sn

可知：a1=1a2=a1+2.an=an-1 + n等式两边分别相加：a1+a2+a3+...+an-1+an = 1 + a1 +2 +a2 +3 +...+an-1+n得：an = 1+2+3+...+n= n(n+1)/2= [n^2+n]/2Sn=[1^2 + 1]/2 + [2^2 + 2]/2+ ...+[n^2 + n]/2=[(1^2 + 2^2 + ....