1*2+2*3+3*4+...+n(n+1)=

因为 k*(k+1) = k² + k
所以 1*2 + 2*3 + 3*4 + ... + n*(n+1)
= (1²+1) + (2²+2) + (3²+3) + ... + (n²+n)
= (1²+2²+3²+...+n²) + (1+2+3+...+n)
= n(n+1)(2n+1)/6 + n(n+1)/2
= [n(n+1)/6] * (2n+1+3)
= n(n+1)(n+2)/3

1*2+2*3+3*4+...n*(n+1)
=1(1+1)+2(2+1)+3(3+1)+···+n(n+1)
=1²+1+2²+2+3²+3+····+n²+n
=(1+2+3+····+n)+(1²+2²+3²+···n²)
=(1+n)n/2+n(n+1)(2n+1)/6
=n(n+1)/2[1+(2n+1)/3]
=n(n+1)(n+2)/3
此题应用的两个常用的求和公式为：
1+2+3+···+n=(1+n)n/2
1²+2²+3²+···n²=n(n+1)(2n+1)/6
这种题目常用加减乘除的结合律，交换律，分配律以及拆分和拼凑来计算，找到其规律，进而得出一般结论。

因为 n*(n+1) = n² + n
所以 1*2 + 2*3 + 3*4 + ... + n*(n+1)
= (1²+1) + (2²+2) + (3²+3) + ... + (n²+n)
= (1²+2²+3²+...+n²) + (1+2+3+...+n)
= n(n+1)(2n+1)/6 + n(n+1)/2
= [n(n+1)/6] * (2n+1+3)
= n(n+1)(n+2)/3

(1*1+2*2+...n*n)+(1+2+...+n)=n*(n+1)*(2n+1)/6+n*(n+1)/2

解:
设第k项为ak
ak=k(k+1)=k²+k
1×2+2×3+...+n×(n+1)
=(1²+2²+3²+...+n²)+(1+2+3+...+n)
=n(n+1)(2n+1)/6+n(n+1)/2
=n(n+1)(n+2)/3