分类：教育/科学-学习帮助
问题：数学求证1×2+2×3+3×4+…+n(n+1)=1/3n(n+ 
求证
1×2+2×3+3×4+…+n(n+1)=1/3 n(n+1)(n+2)
回答：证明:数学归纳法
n=1时
1×2 =(1/3)×1 ×2×3成立
设n=k时,成立,即
Sk=1×2+2×3+3×4+…+k(k+1)=1/3 k(k+1)(k+2)
则n=k+1时
S(k+1)=1×2+2×3+3×4+…+k(k+1)+(k+1)(k+2)
=Sk+(k+1)(k+2)
=1/3 k(k+1)(k+2)+(k+1)(k+2)
提取公因式(1/3)(k+1)(k+2)得
S(k+1)=(1/3)(k+1)(k+2)(k +3)
      =(1/3)(k+1)[(k+1)+1][(k+1)+2]
即n=k+1时也成立
 
所以1×2+2×3+3×4+…+n(n+1)=1/3 n(n+1)(n+2)