5.数列{an}\{a_{n}\}{an}满足an+1=2an−14an+2{a}_{n+1}=\frac{2{a}_{n}-1}{4{a}_{n}+2}an+1=4an+22an−1,且a1=1a_{1}=1a1=1,则数列{an}\{a_{n}\}{an}的前2024项的和S2024=(S_{2024}=(S2024=( )))