1+2+ 3+…+（n+1)=s
(n+1)+n+……+3+2+1=s两式相加得 (n+1+1)+(n+2)+(n-1+3)+……+（2+n)+(1+n+1)=2s即2s=(n+2)+(n+2)+……+(n+2) (共有n+1个）2s=(n+1)(n+2)∴1+2+ 3+…+（n+1)=s=(n+1)(n+2)/2лл��(^_^)������ô��...