The coefficient of $$x^{50}$$ in the binomial expansion of $$(1+x)^{1000} + x(1+x)^{999} + x^2(1+x)^{998} + \ldots + x^{1000}$$ is:

We have to find the coefficient of $$x^{50}$$ in the finite sum

$$S(x)= (1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+\,\ldots\,+x^{1000}.$$

The general (that is, the $$k$$-th) term of this sum is obtained by taking the exponent of $$x$$ equal to $$k$$ and the exponent of $$(1+x)$$ equal to $$1000-k$$. Thus the $$k$$-th term is

$$T_k \;=\;x^{k}\,(1+x)^{1000-k},\qquad k=0,1,2,\ldots,1000.$$

Inside this term the binomial expansion formula

$$ (1+x)^n \;=\;\sum_{r=0}^{n} \binom{n}{r}\,x^{\,r} $$

tells us that the power of $$x$$ that finally appears is $$k+r$$, and its coefficient there is $$\binom{1000-k}{r}.$$

We want the overall power to be exactly $$50$$, so we impose

$$k+r = 50.$$

Since $$k$$ runs from $$0$$ up to $$1000$$, only those $$k$$ with $$k\le 50$$ can possibly satisfy this equality (otherwise $$k+r$$ would exceed $$50$$ because $$r\ge 0$$). Therefore we let

$$k = 0,1,2,\ldots,50.$$

For a fixed $$k$$ in this range the corresponding $$r$$ is

$$r = 50-k,$$ so the contribution of the $$k$$-th term to the coefficient of $$x^{50}$$ is

$$\binom{1000-k}{\,50-k\,}.$$

Hence the required coefficient, call it $$C,$$ is the sum

$$ C \;=\;\sum_{k=0}^{50}\binom{1000-k}{\,50-k\,}. $$

It is convenient to reverse the index so that the upper number inside the binomial sign increases with the index. Put

$$r = 50-k \quad\Longrightarrow\quad k = 50-r.$$

When $$k$$ goes from $$0$$ to $$50$$, the new index $$r$$ goes from $$50$$ down to $$0$$, i.e. from $$0$$ to $$50$$ after re-ordering. Substituting $$k=50-r$$ inside the binomial coefficient gives

$$ \binom{1000-(50-r)}{\,r\,}\;=\;\binom{950+r}{\,r\,}. $$

Therefore

$$ C \;=\;\sum_{r=0}^{50}\binom{950+r}{\,r\,}. $$

Now we invoke a standard “hockey-stick” identity of binomial coefficients:

$$ \sum_{r=0}^{n}\binom{m+r}{\,r\,}\;=\;\binom{m+n+1}{\,n\,}. $$

Here $$m=950$$ and $$n=50,$$ so

$$ C \;=\;\binom{950+50+1}{\,50\,}\;=\;\binom{1001}{\,50\,}. $$

Writing this binomial coefficient in factorial form (using $$\displaystyle\binom{N}{k}=\frac{N!}{k!\,(N-k)!}$$) we obtain

$$ C \;=\;\frac{1001!}{50!\,951!}. $$

This expression exactly matches Option D in the list provided.

Hence, the correct answer is Option D.