Question

x(x+1)分之一加(x+1)(x+2)分之一…一直加到(x+2005)(x+2006)分之一 等于多少?

Answer 1

1/(x+n)(x+n+1)=1/(x+n)-1/(x+n+1)

裂项相消

原式=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-...-1/(x+2005)+1/(x+2005)

-1/(x+2006)

=1/x-1/(x+2006)

=2006/x(x+2006)