Question

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

Answer 1

3*[1*2+2*3+3*4+...n*(n+1)]
根据中括号内的通项公式得
=3*[1(1+1)+2(2+1)+3(3+1)+···+n(n+1)]
=3*[1²+1+2²+2+3²+3+····+n²+n]
移项得
=3*[(1+2+3+····+n)+(1²+2²+3²+···n²)]
由数列求和公式得
=3*[(1+n)n/2+n(n+1)(2n+1)/6]
=3*[n(n+1)/2[1+(2n+1)/3]
=3*[n(n+1)(n+2)/3]
=n(n+1)(n+2)
注：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

Answer 2

解3*[1*2+2*3+3*4+...n*(n+1)]
=3*[1*(1+1)+2*(2+1)+3*(3+1)+...n*(n+1)]
=3*[(1^2+1)+(2^2+2)+(3^2+2)+...+(n^2+n)]
=3*[(1^2+2^2+3^2+n^2)+(1+2+3+....+n)]
=3*[n(n+1)(2n+1)/6+n(n+1)/2]
=3n(n+1)[(2n+1)/6+1/2]
=3n(n+1)(2n+4)/6
=n(n+1)(n+2)

Answer 3

Answer 4

解：原式=3*∑（k^2+k)=3(∑k^2+∑k)（k=1,2，……，n)=3*[n(n+1)(2n+1)/6+n(n+1)/2)]=n(n+1)(n+2)。供参考。