1/(1×4),1/(4×7),...,1/[(3n-2)×(3n+1)]的前几项的和为
1/(1×4)+1/(4×7)+...+1/[(3n-2)×(3n+1)]
=1/3×(1-1/4)+1/3×(1/4-1/7)+……+1/3×[1/(3n-2)-1/(3n+1)]
=1/3×[1-1/4+1/4-1/7+……+1/(3n-2)-1/(3n+1)]
=1/3×[1-1/(3n+1)]
=1/3×(3n)/(3n+1)
=n/(3n+1)
1/(1×4),1/(4×7),...,1/[(3n-2)×(3n+1)]的前几项的和为
1/(1×4)+1/(4×7)+...+1/[(3n-2)×(3n+1)]
=1/3×(1-1/4)+1/3×(1/4-1/7)+……+1/3×[1/(3n-2)-1/(3n+1)]
=1/3×[1-1/4+1/4-1/7+……+1/(3n-2)-1/(3n+1)]
=1/3×[1-1/(3n+1)]
=1/3×(3n)/(3n+1)
=n/(3n+1)