14.数列{an}\{a_{n}\}{an}满足an=2n+1a_{n}=2n+1an=2n+1,且前nnn项和为SnS_{n}Sn,数列{bn}\{b_{n}\}{bn}满足bn=Sn+16n−2b_{n}=\frac{S_n+16}{n}-2bn=nSn+16−2,则∣b1−b2∣+∣b2−b3∣+⋯+∣b15−b16∣\vert b_{1}-b_{2}\vert +\vert b_{2}-b_{3}\vert +\dotsb +\vert b_{15}-b_{16}\vert∣b1−b2∣+∣b2−b3∣+⋯+∣b15−b16∣为((( )))