5.已知SnS_{n}Sn是等比数列{an}\{a_{n}\}{an}的前nnn项和,且Sn=2n+1+a{S}_{n}={2}^{n+1}+aSn=2n+1+a,则a1a2+a2a3+⋯+a10a11=(a_{1}a_{2}+a_{2}a_{3}+\dotsb +a_{10}a_{11}=(a1a2+a2a3+⋯+a10a11=( )))