奥数题1⼀1×2×3+1⼀2×3×4+……+1⼀98×99×100

裂项求和的方法求解前九十八项的和

首先得清楚1/(a-1)a=1/(a-1)-1/a;
则1/(a-1)a(a+1)=1/2a[1/(a-1)-1/(a+1)]=1/2[1/(a-1)-2/a+1/(a+1)]

原式=1/2[(1-1+1/3)+(1/2-2/3+1/4)+...(1/98-2/99+1/100)
=1/2(1-1+1/2+1/98-2/99+1/100)
=4949/19800

1/(a-1)a(a+1)=1/2a[1/(a-1)-1/(a+1)]=1/2[1/(a-1)-2/a+1/(a+1)]
如1/98×99×100=1/2(1/98-2/99+1/100)
所以原式=1/2[(1-1+1/3)+(1/2-2/3+1/4)+...
=1/2(1-1+1/2+1/99-2/99+1/100)
=4949/19800

先做1/1×2+1/2×3+……+1/98×99这题，再做上边这题

1/1×2×3+1/2×3×4+……+1/98×99×100
=1/2*（1/1*2-1/2*3)+1/2*(1/2*3-1/3*4)+......+1/2*(1/98*99-1/99*100)
=1/2*(1/1*2-1/99*100)
=1/4-1/19800
=4949/19800