Question

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

• A. $\frac{(1000)!}{(50)! ( 950)!}$
• B. $\frac{(1000)!}{(49)! ( 951)!}$
• C. $\frac{(1001)!}{(51)! ( 950)!}$
• D. $\frac{(1001)!}{(50)! ( 951)!}$

Answer 1

The Correct Option is D

To solve the problem of finding the coefficient of $x^{50}$ in the given expression, we start by analyzing the structure of the expression:

The given series is:

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

Observe that each term in the series is of the form $x^k (1 + x)^{1000-k}$ where $k$ ranges from 0 to 1000.

We need to find the coefficient of $x^{50}$ in this expansion.

The general term of the form in the series is:

$T_k = x^k \cdot (1 + x)^{1000-k}$

The term $(1 + x)^{1000-k}$ can be expanded using the binomial theorem:

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

So, each term will be:

$T_k = x^k \cdot \sum_{r=0}^{1000-k} \binom{1000-k}{r} x^r = \sum_{r=0}^{1000-k} \binom{1000-k}{r} x^{r+k}$

We need the total power of $x$ to be 50, that is:

$r + k = 50$

Thus:

$r = 50 - k$

Substitute this in the binomial coefficient:

$T_k = \binom{1000-k}{50-k} x^{50}$

Now, sum up the coefficients of $x^{50}$ from all terms:

$\sum_{k=0}^{50} \binom{1000-k}{50-k}$

Using the hockey-stick identity of combinatorics, we have:

$\sum_{i=0}^{r} \binom{n+i}{i} = \binom{n+r+1}{r}$

Apply it to get the coefficient:

$\sum_{k=0}^{50} \binom{1000-k}{50-k} = \binom{1001}{50}$

The formula for the binomial coefficient is:

$\binom{n}{r} = \frac{n!}{r! (n-r)!}$

Therefore, the coefficient of $x^{50}$ is:

$\frac{1001!}{50! \cdot 951!}$

Thus, the correct answer is:

$\frac{(1001)!}{(50)! (951)!}$, which matches one of the provided options.