The sum of the coefficients of $$x^{499} \text{ and }x^{500} \text{ in } (1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+....+x^{1000} \text{ is: }$$

Write the given expression as a finite sum:

$$S(x)=\sum_{k=0}^{1000} x^{k}(1+x)^{1000-k}$$

Our goal is to find the coefficients of $$x^{499}$$ and $$x^{500}$$ in $$S(x)$$ and then add them.

Step 1: Coefficient of $$x^{n}$$ in a single term.
For a fixed $$k$$, expand $$(1+x)^{1000-k}$$ using the Binomial Theorem:

$$x^{k}(1+x)^{1000-k}=x^{k}\sum_{r=0}^{1000-k} \binom{1000-k}{r}x^{r} =\sum_{r=0}^{1000-k} \binom{1000-k}{r}x^{k+r}$$

Hence the coefficient of $$x^{n}$$ coming from the term with index $$k$$ equals
$$\binom{1000-k}{\,n-k\,}\quad\text{provided }0\le n-k\le 1000-k$$

Step 2: Total coefficient of $$x^{n}$$ in $$S(x)$$.
Sum the contributions from all $$k$$ that can reach power $$n$$ (namely $$k=0$$ to $$k=n$$):

$$c_n=\sum_{k=0}^{n}\binom{1000-k}{\,n-k\,}$$

Step 3: Re-index the sum.
Put $$j=n-k\;(\Rightarrow k=n-j)$$. Then $$j$$ runs from $$0$$ to $$n$$ and

$$c_n=\sum_{j=0}^{n}\binom{(1000-n)+j}{\,j\,}$$

Step 4: Use the standard identity.
For any non-negative integers $$m,n$$,

$$\sum_{j=0}^{n}\binom{m+j}{j}=\binom{m+n+1}{n}\quad -(1)$$

Here $$m=1000-n$$, so from $$(1)$$ we get

$$c_n=\binom{1000-n+n+1}{n}=\binom{1001}{n}$$

Step 5: Evaluate the required coefficients.

Coefficient of $$x^{499}$$: $$c_{499}=\binom{1001}{499}$$
Coefficient of $$x^{500}$$: $$c_{500}=\binom{1001}{500}$$

Step 6: Add the two coefficients.
Using Pascal’s rule $$\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}$$ with $$n=1001,\;r=499$$:

$$c_{499}+c_{500} =\binom{1001}{499}+\binom{1001}{500} =\binom{1002}{500}$$

Final Answer: $$\displaystyle \binom{1002}{500}$$ (Option D).