jmc

We know that by definition, $i^2=-1$, so $i^4=(-1{)}^2=1.$ More generally, for all integers k, $i^{4k}=(i^4{)}^k=1^k=1$. This means that $i^{11}+i^{16}+i^{21}+i^{26}+i^{31}=i^8(i^3)+i^{16}(1)+i^{20}(i)+i^{24}(i^2)+i^{28}(i^3)=i^3+1+i+i^2+i^3$. Since $i^3=-i$, we can simplify this to get our final result: $i^{11}+i^{16}+i^{21}+i^{26}+i^{31}=-i+1+i-1-i=\boxed{-i}.$