Coefficient of ${\mathrm{x}}^{2016}\text{ in }{\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3+{\mathrm{x}}^4\right)}^{1001}(1-\mathrm{x}{)}^{1002}\left(1+7{\mathrm{x}}^{14}\right)$ is

In the given expression, consider ${\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3+{\mathrm{x}}^4\right)}^{1001}$

The expression inside the bracket $1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$, this is from sum of first five terms of geometry progression having first term 1 and the common ratio $x$d

Hence, the given expression is 

$\begin{array}{rcl}{\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3+{\mathrm{x}}^4\right)}^{1001}(1-\mathrm{x}{)}^{1002}\left(1+7{\mathrm{x}}^{14}\right)& =& {\left(1-{\mathrm{x}}^5\right)}^{1001}(1-\mathrm{x}{)}^{-1001}(1-\mathrm{x}{)}^{1002}\left(1+7{\mathrm{x}}^{14}\right)\\ & =& {\left(1-{\mathrm{x}}^5\right)}^{1001}(1-\mathrm{x})\left(1+7{\mathrm{x}}^{14}\right)\\ & =& \left(1+7x^{14}-x-7x^{15}\right){\left(1-x^5\right)}^{1001}\end{array}$

To get the coefficieint of $x^{2016}$, 

Find the values of $r$ such that 

1. $5r=2016$
2. $5r+1=2016$
3. $5r+14=2016$
4. $5r+15=2016$

because $5r+1=2016$ only will get $r$value as integer.

find the coefficient of 404th term in the expansion of $\begin{array}{rcl} & & {\left(1-x^5\right)}^{1001}\end{array}$ and then multiply with $-1$

hence the required coefficient is $\boxed{{}^{1001}C_{403}}$