1x2²+2x3²+3x4²+.....+19x20²

Question

求解

Answer 1

1×2²+2×3²+3×4²+……18+19²×+19×20²
=（2-1）×2²+（3-1）×3²+（4-1）×4²+……（19-1）19²×+（20-1）×20²
=（2³+...+n ³）-（2²+3²+4²+……19²×+20²）
自然数平方和=1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6
自然数立方和= 1³+ 2³+...+n ³= n²(n+1)²/4
带入得41230本回答被提问者采纳

Answer 2

Answer 3

∑n(n+1)²=∑n(n²+2n+1)=∑n³+2n²+n
=[n(n+1)/2]²+2×[n(n+1)(2n+1)/6]+[n(n+1)/2]
=[n(n+1)/2]²+[n(n+1)(2n+1)/3]+[n(n+1)/2]
=190²+(19×20×39/3)+190
=190²+19×20×13+190
=41230