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**Hint:**Try to write the expression in expanded form without summation. Notice that on the left side there is a geometric sequence and on the right side there is a polynomial, whose term ${a_{17}}{x^{17}}$ is of most importance to us. Find the sum of geometric progression using the formula. Now using the binomial theorem on the sum of a geometric sequence, figure out the coefficient of ${x^{17}}$.

**Complete step-by-step answer:**

Let’s first analyse the given equation in the question and rewrite it in expanded form as:

$ \Rightarrow {\left( {1 + x} \right)^{2016}} + x{\left( {1 + x} \right)^{2015}} + {x^2}{\left( {1 + x} \right)^{2014}} + ..... + {x^{2016}} = {a_0}{x^0} + {a_1}{x^1} + {a_2}{x^2} + {a_3}{x^3}.......... + {a_{2016}}{x^{2016}}$

So, here on the left-hand side (LHS), we have a series of geometric progression with a common ratio of $\dfrac{x}{{\left( {x + 1} \right)}}$ and on the right-hand side (RHS), we have a polynomial with a maximum power of ${x^{2016}}$.

And using this equation, we have to find the value of ${a^{17}}$, i.e. the coefficient of ${x^{17}}$ from the polynomial on RHS.

Let’s first find the sum of this geometric progression on the LHS

As we know, a geometric sequence is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called the common ratio which is denoted by r. The common ratio is obtained by dividing any term by the preceding term.

And for a sequence with the number of elements as n, first term as a, the sum of a geometric progression is defined as: $S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$

So, we can use this expression to find the sum of all terms in the LHS of the equation. We have here $a = {\left( {1 + x} \right)^{2016}}$ , $r = \dfrac{x}{{1 + x}}$ and number of terms as $2017$

$ \Rightarrow LHS = \dfrac{{{{\left( {1 + x} \right)}^{2016}}\left[ {{{\left( {\dfrac{x}{{1 + x}}} \right)}^{2017}} - 1} \right]}}{{\left( {\dfrac{x}{{1 + x}} - 1} \right)}} = \dfrac{{{{\left( {1 + x} \right)}^{2016}} \times \dfrac{{\left[ {{x^{2017}} - {{\left( {1 + x} \right)}^{2017}}} \right]}}{{{{\left( {1 + x} \right)}^{2017}}}}}}{{\dfrac{{\left( {x - 1 - x} \right)}}{{1 + x}}}} = - \left[ {{x^{2017}} - {{\left( {1 + x} \right)}^{2017}}} \right]$

$ \Rightarrow LHS = {\left( {1 + x} \right)^{2017}} - {x^{2017}}$

So, our previous long equation became:

$ \Rightarrow {\left( {1 + x} \right)^{2017}} - {x^{2017}} = {a_0}{x^0} + {a_1}{x^1} + {a_2}{x^2} + {a_3}{x^3}.......... + {a_{2016}}{x^{2016}}$

Now, we just need to find the coefficient of the ${x^{17}}$ from the expression ${\left( {1 + x} \right)^{2017}} - {x^{2017}}$

But since ${x^{2017}}$can not cause any change to the coefficient of ${x^{17}}$, we need to find the required answer from the expansion of${\left( {1 + x} \right)^{2017}}$.

Using the Binomial expansion to solve this, will be the best choice here.

According to the Binomial expansion:${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{x^{n - r}}{y^r}} $ and for the expression here, we have $n = 2017,y = 1$

$ \Rightarrow {\left( {1 + x} \right)^{2017}} = {}^{2017}{C_0}{x^0} + {}^{2017}{C_1}{x^1} + {}^{2017}{C_2}{x^2}........... + {}^{2017}{C_{2017}}{x^{2017}}$

Therefore, from the above expression, we can conclude that the coefficient of ${x^{17}}$ will is ${}^{2017}{C_{17}}$.

Hence, ${a_{17}} = {}^{2017}{C_{17}} = \dfrac{{2017!}}{{17!\left( {2017 - 17} \right)!}} = \dfrac{{2017!}}{{17!2000!}}$

**So, the correct answer is “Option D”.**

**Note:**Try to go step by step with the solution. Using the sum of geometric progression and binomial theorem was the most crucial part of the solution. The use of binomial expansion will always help you in figuring out the coefficient of a specific term without solving.In above step the total number of terms in expansion $(x+y)^n$ is given by formula $(n+1)$.

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