jmc

We see that $a_n=4n^3+6n^2+4n+1=(n^4+4n^3+6n^2+4n+1)-n^4=(n+1{)}^4-n^4,$ so $$a_8+a_9+a_{10}+\cdots +a_{23}=(9^4-8^4)+(10^4-9^4)+(11^4-10^4)+\cdots +(24^4-23^4)=24^4-8^4=\boxed{327680}.$$