jmc

Let $$S=(1+x{)}^{1000}+2x(1+x{)}^{999}+\cdots +1000x^{999}(1+x)+1001x^{1000}$$Then $$\begin{array}{rl}xS& =x(1+x{)}^{1000}+2x^2(1+x{)}^{999}+\cdots +1000x^{1000}(1+x)+1001x^{1001},\\ (1+x)S& =(1+x{)}^{1001}+2x(1+x{)}^{1000}+\cdots +1000x^{999}(1+x{)}^2+1001x^{1000}(1+x).\end{array}$$Subtracting these equations, we get $$S=(1+x{)}^{1001}+x(1+x{)}^{1000}+\cdots +x^{999}(1+x{)}^2+x^{1000}(1+x)-1001x^{1001}.$$Then $$\begin{array}{rl}xS& =x(1+x{)}^{1001}+x^2(1+x{)}^{1000}+\cdots +x^{1000}(1+x{)}^2+x^{1001}(1+x)-1001x^{1002},\\ (1+x)S& =(1+x{)}^{1002}+x(1+x{)}^{1001}+\cdots +x^{999}(1+x{)}^3+x^{1000}(1+x{)}^2-1001x^{1001}(1+x).\end{array}$$Subtracting these equations, we get $$S=(1+x{)}^{1002}-1002x^{1001}(1+x)+1001x^{1002}.$$By the Binomial Theorem, the coefficient of $x^{50}$ is then $\left(\genfrac{}{}{0ex}{}{1002}{50}\right).$ The final answer is $1002+50=\boxed{1052}.$