1×2+2×3+……+n(n+1) 等于多少

把通项拆分成n^2和n，再求和。结果是n(n+1)(2n+1)/6+n(n+1)/2

裂项法
n(n+1)=1/3[n(n+1)(n+2)-(n-1)n(n+1)
所以原式=1/3[1*2*3-0*1*2+2*3*4-1*2*3+3*4*5-2*3*4+……+n(n+1)(n+2)-(n-1)n(n+1)]
=1/3[-0*1*2+n(n+1)(n+2)]
=1/3[n(n+1)(n+2)]

原式=1^2+2^2+3^2+…n^2+1+2+3+…+n
=n*(n+1)*(2n+1)/6+n*(n+1)/2
=n*(n+1)*(n+2)/3