因为数列{a_n}的前n项和为Sn=lg[

1

10

(1+n)]，

所以a₁₀+a₁₁+a₁₂+…+a₉₉

=S₉₉-S₉

=lg[

1

10

(1+99)]- lg[

1

10

(1+9)]

=1

故答案为1．