设数列{an}:1,﹣2,﹣2,3,3,3,﹣4,﹣4,﹣4,﹣4,…, (−1)k-1k,⋯,(−1)k-1k︷k个 ,…,即当 (k−1)k2 <n≤ (k+1)k2 (k∈N*)时, an=(−1)k−1k .记Sn=a1+a2+…+an(n∈N∗).对于l∈N∗,定义集合Pl=﹛n|Sn为an的整数倍,n∈N∗,且1≤n≤l}
—Yes, it's a major way of _______ for them.