Question

用裂项法求值1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)1/1x3 +

Answer 1

1/n(n+1)(n+2)=[1/n(n+1)-1/(n+1)(n+2)]/2

1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)

=[和1/n(n+1)-和1/(n+1)(n+2)]/2

=[1-1/(n+1)-1/2+1/(n+2)]/2

=1/4-1/2(n+1)(n+2)

1/n(n+2)=[1/n-1/(n+2)]/2

1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)

=[1+1/2-1/(n+1)-1/(n+2)]/2

=3/4-1/2(n+1)-1/2(n+2)