In expansion of $(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$ . Find the coefficient of $x^{28}$ I am not able to apply binomial theorem here
$(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$ 
Please help me to find the coefficient of$ x^{28}$
Any help will be appreciated.

It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.

In order to determine the coefficient of $x^{28}$  we obtain
  \begin{align*}
\color{blue}{[x^{28}]}&\color{blue}{(1+x+\cdots+x^{27})(1+x+\cdots+x^{14})^2}\\
&=[x^{28}]\frac{1-x^{28}}{1-x}\cdot\frac{(1-x^{15})^2}{(1-x)^2}\tag{1}\\
&=[x^{28}]\frac{(1-x^{28})(1-2x^{15}+x^{30})}{(1-x)^3}\\
&=[x^{28}](1-2x^{15}-x^{28})\sum_{j=0}^\infty\binom{-3}{j}(-x)^j\tag{2}\\
&=\left([x^{28}]-2[x^{13}]-[x^0]\right)\sum_{j=0}^\infty\binom{j+2}{j}x^j\tag{3}\\
&=\binom{30}{2}-2\binom{15}{2}-\binom{2}{0}\tag{4}\\
&=435-210-1\\
&\color{blue}{\,\,=224}
\end{align*}

Comment:


*

*In (1) we apply the finite geometric series formula.

*In (2) we expand the numerator $(1-x^{28})(1-x^{15})^2$ and skip all terms with exponent greater $28$ since they don't contribute to the coefficient of $x^{28}$. We also apply the binomial series expansion

*In (3) we use the linearity of the coefficient of operator and apply the rule
$[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also apply the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q
\end{align*}

*In (4) we select the coefficients accordingly.
