已知数列$\{a_n\}$满足$a_{n+1}=\left[\dfrac{a_n}2\right]$，且$a_1=34567$，则其前$n$项和的最大值为_______．
分析与解 考虑到$$34567_{(10)}=1000011100000111_{(2)},$$而每次除以$2$取整相当于抹掉二进制数的最后一位，于是所求的最大值\[\begin{split} M&=1000011100000111_{(2)}+100001110000011_{(2)}+\cdots+10_{(2)} +1_{(2)}\\&=\left(\underbrace{11\cdots 1}_{16}+\underbrace{11\cdots 1}_{10}+\cdots +1\right)_{(2)}\\&=\left(\underbrace{11\cdots 1}_{16}+1+\underbrace{11\cdots 1}_{10 }+1+\cdots +1+1\right)_{(2)}-7\\&=10000111000001110_{(2)}-7_{(10)}\\&=2a_1-7\\
&=69127.\end{split}\]