For $$x \in R$$, $$x \neq -1$$, if $$(1+x)^{2016} + x(1+x)^{2015} + x^2(1+x)^{2014} + \ldots + x^{2016} = \sum_{i=0}^{2016} a_i x_i$$, then $$a_{17}$$ is equal to

We start with the given finite sum

$$S \;=\; (1+x)^{2016} \;+\; x(1+x)^{2015} \;+\; x^{2}(1+x)^{2014} \;+\; \ldots \;+\; x^{2016},$$

defined for all real $$x$$ with $$x\neq -1$$. It is convenient to write the same expression with an index so that every term follows the same pattern:

$$S \;=\; \sum_{k=0}^{2016} x^{k}\,(1+x)^{2016-k}.$$

At this point we recall the algebraic identity for the difference of two powers:

$$a^{n+1}-b^{n+1} \;=\; (a-b)\sum_{k=0}^{n} a^{\,n-k}\,b^{k}.$$

This identity is valid for all real or complex numbers $$a,b$$ and every non-negative integer $$n$$. We now apply it with the specific choices

$$a = 1+x,\qquad b = x,\qquad n = 2016.$$

Substituting these values in the general formula, we obtain

$$(1+x)^{2017} \;-\; x^{2017} \;=\; \bigl((1+x)-x\bigr)\, \sum_{k=0}^{2016} (1+x)^{2016-k}\,x^{k}.$$

Inside the large parentheses we plainly have

$$(1+x)-x \;=\; 1.$$

Hence the right-hand side simplifies to

$$(1+x)^{2017} - x^{2017} \;=\; 1 \times \sum_{k=0}^{2016} (1+x)^{2016-k}\,x^{k} \;=\; \sum_{k=0}^{2016} x^{k}(1+x)^{2016-k}.$$

The last sum is exactly our original $$S$$, therefore we have established the compact form

$$S \;=\; (1+x)^{2017} \;-\; x^{2017}.$$

Our task is to find $$a_{17}$$, the coefficient of $$x^{17}$$ when $$S$$ is expanded as a polynomial in $$x$$. The representation we have just derived makes this easy:

• The term $$(1+x)^{2017}$$ contains a contribution $$\binom{2017}{17}x^{17}$$.
• The term $$-\,x^{2017}$$ does not contain any $$x^{17}$$ part, because its only term is $$-x^{2017}$$.

Therefore the entire coefficient of $$x^{17}$$ in $$S$$ is simply

$$\binom{2017}{17} \;=\; \frac{2017!}{17!\,2000!}.$$

Thus

$$a_{17} \;=\; \frac{2017!}{17!\,2000!}.$$

Comparing with the options, we see that this value corresponds to Option A.

Hence, the correct answer is Option A.