等差数列求和问题1+2+3+······+n=n(n+1)/2.方法二：s=1+2+3+······+n…①则s=n+······+3+2+1…②①+②得2s=(n+1)+······+（n+1)+（n+1)+（n+1)=n(n+1)所以s=n(n+1)/2.
1+2+3.+N等于(1/2)n(n＋1)
