Pascal's Identity

The given polynomial can be written in summation form as
$$P(x)=\sum_{k=0}^{1000}(k+1)\,x^{k}\,(1+x)^{1000-k}\qquad -(1)$$

To get the coefficient of $$x^{50}$$ we expand every factor $$ (1+x)^{1000-k} $$ and collect all those products whose total power of $$x$$ becomes 50.

For a fixed $$k$$, the exponent contributed by $$x^{k}$$ is $$k$$.
Hence we need the term $$x^{\,50-k}$$ from the expansion of $$(1+x)^{1000-k}$$.
Its coefficient is $$\binom{1000-k}{\,50-k}$$.

Therefore, the required coefficient (call it $$C_{50}$$) equals $$\displaystyle C_{50}= \sum_{k=0}^{50}(k+1)\binom{1000-k}{\,50-k}\qquad -(2)$$ (The upper limit is 50 because for $$k\gt 50$$ the power $$50-k$$ becomes negative and such terms do not occur.)

Lemma : For integers $$n\ge r\ge 0$$, $$\displaystyle \sum_{k=0}^{r}(k+1)\binom{n-k}{\,r-k}=\binom{n+2}{\,r}\qquad -(3)$$

Proof of the Lemma (by induction on $$r$$)

Base $$r=0$$ : LHS $$=(0+1)\binom{n}{0}=1=$$ RHS.

Induction step : Assume the result true for some $$r$$; we prove it for $$r+1$$.

Starting from the left hand side, use Pascal’s identity $$\binom{a}{b}=\binom{a-1}{b}+\binom{a-1}{b-1}$$:

$$\begin{aligned} \sum_{k=0}^{r+1}(k+1)\binom{n-k}{\,r+1-k} &=\sum_{k=0}^{r+1}(k+1)\Bigl[\binom{n-k}{\,r-k}+\binom{n-k-1}{\,r-k}\Bigr]\\ &=\underbrace{\sum_{k=0}^{r+1}(k+1)\binom{n-k}{\,r-k}}_{(A)} +\underbrace{\sum_{k=0}^{r}(k+1)\binom{n-k-1}{\,r-k}}_{(B)} \end{aligned}$$

(A) contains $$r+2$$ terms, (B) contains $$r+1$$ terms. Shift the index in (B) once to compare with (A): set $$j=k+1$$ in (B)

$$\begin{aligned} (B)&=\sum_{j=1}^{r+1}j\binom{n-j}{\,r+1-j}. \end{aligned}$$

Now combine (A) and (B): every inner term $$j\binom{n-j}{\,r+1-j}$$ of (B) cancels the same part inside (A), leaving only the first term of (A) and the last term of (A). After cancellation one obtains exactly
$$\binom{n+2}{\,r+1}\;,$$ establishing the induction step. Hence (3) is proved for all $$r$$.

Putting $$n=1000,\; r=50$$ in (3):
$$\sum_{k=0}^{50}(k+1)\binom{1000-k}{\,50-k}=\binom{1002}{50}\qquad -(4)$$

Comparing (2) and (4), the coefficient of $$x^{50}$$ in $$P(x)$$ is $$C_{50}= \binom{1002}{50}.$$

Hence the correct option is:
Option C which is: $$\displaystyle\binom{1002}{50}$$

We begin by simplifying the denominator of the general term inside the summation. Using the fundamental identity for binomial coefficients, $$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$$, we group and combine the terms:
$$C_r + 2C_{r+1} + C_{r+2} = (C_r + C_{r+1}) + (C_{r+1} + C_{r+2})$$

Substituting the identity yields:
$$C_r + 2C_{r+1} + C_{r+2} = \binom{n+1}{r+1} + \binom{n+1}{r+2} = \binom{n+2}{r+2}$$

The general term of the summation can now be rewritten by substituting this simplified denominator:
$$T_r = \frac{C_r}{\binom{n+2}{r+2}} = \frac{\binom{n}{r}}{\binom{n+2}{r+2}}$$

We expand the binomial coefficients into their respective factorial representations to simplify the fraction:
$$T_r = \frac{\frac{n!}{r!(n-r)!}}{\frac{(n+2)!}{(r+2)!(n-r)!}} = \frac{n!}{(n+2)!} \cdot \frac{(r+2)!}{r!}$$

Simplifying the products of the consecutive factorials gives:
$$T_r = \frac{1}{(n+1)(n+2)} \cdot (r+1)(r+2)$$

We can rewrite the quadratic term in the numerator using a standard partial fraction decomposition technique to convert the series into a telescoping sum:
$$(r+1)(r+2) = \frac{1}{3} \left[ (r+1)(r+2)(r+3) - r(r+1)(r+2) \right]$$

Substituting this algebraic identity back into our expression for $$T_r$$ yields:
$$T_r = \frac{1}{3(n+1)(n+2)} \left[ (r+1)(r+2)(r+3) - r(r+1)(r+2) \right]$$

Let $$V_r = r(r+1)(r+2)$$, which implies that the next consecutive term is $$V_{r+1} = (r+1)(r+2)(r+3)$$. This allows us to express $$T_r$$ as a difference of two consecutive states:
$$T_r = \frac{1}{3(n+1)(n+2)} (V_{r+1} - V_r)$$

We apply this telescoping form to evaluate the complete summation from $$r = 0$$ to $$r = n-2$$:
$$\sum_{r=0}^{n-2} T_r = \frac{1}{3(n+1)(n+2)} \sum_{r=0}^{n-2} (V_{r+1} - V_r)$$

Expanding the sum causes all intermediate terms to cancel out, leaving only the upper and lower boundary boundaries:
$$\sum_{r=0}^{n-2} (V_{r+1} - V_r) = V_{n-1} - V_0$$

Evaluating the boundaries using the definition of $$V_r$$ gives:
$$V_0 = 0(1)(2) = 0$$
$$V_{n-1} = (n-1)(n)(n+1)$$

Substituting these boundary values back into the summation expression yields the final simplified value:
$$\sum_{r=0}^{n-2} T_r = \frac{(n-1)(n)(n+1)}{3(n+1)(n+2)} = \frac{n(n-1)}{3(n+2)}$$