求数列2/(1乘2),2/(2乘3),2/(3乘4),2/(4乘5)...的前n项和
4个回答

令:S=2/(1×2)+2/(2×3)+2/(3×4)+.+2/[n(n+1)]

因此:

S/2=1/(1×2)+1/(2×3)+1/(3×4)+.+1/[n(n+1)]

注意到:

1/(1×2) = (2-1)/(1×2)=1-1/2

1/(2×3) =(3-2)/(2×3)=1/2-1/3

.

1/[n(n+1)]=[(n+1)-n]/[n(n+1)]=1/n - 1/(n+1)

因此:

S/2=1/(1×2)+1/(2×3)+1/(3×4)+.+1/[n(n+1)]

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+.+1/n-1/(n+1)

=1+(-1/2+1/2)+(-1/3+1/3)+(-1/4+1/4)+(-1/5+1/5)+(-1/6+.+[-1/n+1/n]-1/(n+1)

=1-1/(n+1)

=n/(n+1)

S=2n/(n+1)