题目

1*100+2*99+3*98+4*97+5*96+.+50*51=?,要有过程的、、、

提问时间：2020-08-07

答案

原式=1*100+2*(100-1)+3*(100-2)+4*(100-3)+……+50*(100-49) =1*100+2*100-2*1+3*100-3*2+4*100-4*3+……+50*100-50*49 =(1+2+3+4+……+50)*100-(1*2+2*3+3*4+……+49*50) =(1+50)*50/2*100-49*50*51/3 =1275*100-41650 =127500-41650 =85850 附：设S=1*2+2*3+3*4+……n*(n+1) 则3S=1*2*3+2*3*3+3*4*3+4*5*3+……n*(n+1)*3 =1*2(3-0)+2*3(4-1)+3*4(5-2)+……n(n+1)[n+2-(n-1)] =1*2*3+2*3*4-1*2*3+3*4*5-2*3*4+……+n(n+1)(n+2)-(n-1)n(n+1) =n(n+1)(n+2),所以1*2+2*3+3*4+……+49*50=S=n(n+1)(n+2)/3 故S=49*(49+1)*(49+2)/3=41650