Question

1×2分之2+2×3分之2+3×4分之2+...+n×(n+1)分之2等于什么?

要过程

Answer 1

2/(1×2)+2(2×3)+2/(3×4)+...+2/[n(n+1)]
=2{1/(1×2)+1/(2×3)+1/(3×4)+...+1/[n(n+1)]}
=2[1-1/2+1/2-1/3+1/3-1/4+...+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)

Answer 2

1×2分之2+2×3分之2+3×4分之2+...+n×(n+1)分之2
= 2(1-1/2) + 2(1/2-1/3) + 2(1/3-1/4) + ...... + 2{1/n-1/(N+1)}
= 2{1 - 1/(n+1)}
= 2n/(n+1)