<s>[INST] 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.
 [/INST] Express is as a product of three factors $F_1,F_2,F_3$:
$$(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}=(1+\cdots +x^{27})(1+\cdots +x^{14})(1+\cdots +x^{14}).$$
Make up a table of powers of $x$:
$$\begin{array}{c|c|lcr}
m& n & \text{$F_1$} & \text{$F_2$} & \text{$F_3$} \\
\hline
& 1 & 14 & 14 & 0 \\
& \downarrow & \downarrow & \downarrow & \vdots \\
1 & 14 & 27 & 1 & 0 \\
& 1 & 13 & 14 & 1 \\
& \downarrow & \downarrow & \downarrow & \vdots \\
1 &15 & 27 & 0 & 1 \\
& 1 & 12 & 14 & 2 \\
&\downarrow & \downarrow & \downarrow & \vdots \\
2 &15 & 26 & 0 & 2 \\
&1 & 11 & 14 & 3 \\
&\downarrow & \downarrow & \downarrow & \vdots \\
3 & 15 & 25 & 0 & 3 \\
& 1 & 10 & 14 & 4 \\
& \downarrow & \downarrow & \downarrow & \vdots \\
4 & 15 & 24 & 0 & 4 \\
& 1 & 9 & 14 & 5 \\
& \downarrow & \downarrow & \downarrow & \vdots \\
5 & 15 & 23 & 0 & 5 \\
& \cdots & \cdots & \cdots & \cdots \\
& 1 & 0 & 14 & 14 \\
& \downarrow & \downarrow & \downarrow & \vdots \\
14 & 15 & 14 & 0 & 14 \\
\end{array}$$
Hence:
$$1\cdot 14 + 14\cdot 15=224.$$
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