13.已知数列{an}\{a_{n}\}{an}的前nnn项和为SnS_{n}Sn,且满足an={2n−1,n=2k−1,k∈N∗5n+1,n=2k,k∈N∗{a}_{n}=\left\{\begin{array}{l}2n-1,&n=2k-1,k\in {N}^{*}\\ 5n+1,&n=2k,k\in {N}^{*}\end{array}\right.an={2n−1,5n+1,n=2k−1,k∈N∗n=2k,k∈N∗,则S10=(S_{10}=(S10=( )))