答案：41650 一般的,（用*表示相乘） 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), 所以原式=S=n(n+1)(n+2)/3故S=49*(49+1)*(49+2)/3=41650