For two positive integers $a$ and $b$, the Ulam sequence $U(a,b)$ is defined by $U(a,b{)}_1=a$, $U(a,b{)}_2=b$ and for $k>2$, $U(a,b{)}_k$ is the smallest integer greater than $U(a,b{)}_{k-1}$ which can be written in exactly one way as the sum of two distinct previous members of $U(a,b)$.

For example, the sequence $U(1,2)$ begins with
$1$, $2$, $3=1+2$, $4=1+3$, $6=2+4$, $8=2+6$, $11=3+8$;
$5$ does not belong to it because $5=1+4=2+3$ has two representations as the sum of two previous members, likewise $7=1+6=3+4$.

Find $\sum _{n=2}^{10}U(2,2n+1{)}_k$, where $k={10}^{11}$.