计算1/1×2×3 +1/2×3×4 +1/3×4×5 +……+1/48×49×50 怎样算简便

Question

六年级奥数计算1/1×2×3 +1/2×3×4 +1/3×4×5 +……+1/48×49×50 怎样计算简便

Answer 1

不好意思看错题了,,,, 更正一下: 1/(k-1)k(k+1) =1/k(k^2-1) =k/(k^2-1)-1/k =1/2(k-1)+1/2(k+1)-1/k [49>=k>=2] 就按这种思路把每项分解开再求和。 好吧，那就给你算出来吧。 通项 =1/2(k-1)+1/2(k+1)-1/k =[1/(k-1)+1/(k+1)]/2-1/k 原式 =[1/1+1/3]/2-1/2 +[1/2+1/4]/2-1/3 +[1/3+1/5]/2-1/4 +[1/4+1/6]/2-1/5 ... +[1/46+1/48]/2-1/47 +[1/47+1/49]/2-1/48 +[1/48+1/50]/2-1/49 上式各项同乘以2，整体再除以2，其值不变。 ∴原式 ={[1/1+1/3]-2/2 +[1/2+1/4]-2/3 +[1/3+1/5]-2/4 +[1/4+1/6]-2/5 ... +[1/46+1/48]-2/47 +[1/47+1/49]-2/48 +[1/48+1/50]-2/49}/2 ={1/2-1/49+1/50}/2 =(25*49-50+49)/2*49*50 =1224/4900 =306/1225