1个回答
展开全部
解法一:
n(n+1)=⅓×[n(n+1)(n+2)-(n-1)n(n+1)]
1×2+2×3+...+n(n+1)
=⅓×[1×2×3-0×1×2+2×3×4-1×2×3+...+n(n+1)(n+2)-(n-1)n(n+1)]
=⅓n(n+1)(n+2)
解法二:
n(n+1)=n²+n
1×2+2×3+...+n(n+1)
=(1²+2²+...+n²)+(1+2+...+n)
=n(n+1)(2n+1)/6 +n(n+1)/2
=[n(n+1)/6](2n+1+3)
=n(n+1)(2n+4)/6
=⅓n(n+1)(n+2)
n(n+1)=⅓×[n(n+1)(n+2)-(n-1)n(n+1)]
1×2+2×3+...+n(n+1)
=⅓×[1×2×3-0×1×2+2×3×4-1×2×3+...+n(n+1)(n+2)-(n-1)n(n+1)]
=⅓n(n+1)(n+2)
解法二:
n(n+1)=n²+n
1×2+2×3+...+n(n+1)
=(1²+2²+...+n²)+(1+2+...+n)
=n(n+1)(2n+1)/6 +n(n+1)/2
=[n(n+1)/6](2n+1+3)
=n(n+1)(2n+4)/6
=⅓n(n+1)(n+2)
本回答被网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询