<s>[INST] Generating function - What is the coefficient of $x^{26}$? What is the coefficient of $x^{26}$ given the function:
$${\left[ {{{1 + {x^{10}}} \over {{{(1 - x)}^7}}}} \right]^2}$$
My work:
$${\left[ {{{1 + {x^{10}}} \over {{{(1 - x)}^7}}}} \right]^2} = \left( {1 + 2{x^{10}} + {x^{20}}} \right){1 \over {{{(1 - x)}^{14}}}} = \left( {1 + 2{x^{10}} + {x^{20}}} \right)\sum\limits_{n = 0}^\infty  {\left( {\matrix{
   {n + 13}  \cr 
   {13}  \cr 
 } } \right)}  \cdot {x^n}$$
Recitation's notes work:
$$... = \left( {{x^{26}} + 2{x^{16}} + {x^6}} \right)\sum\limits_{n = 0}^\infty  {\left( {\matrix{
   {n + 13}  \cr 
   {13}  \cr 
 } } \right)}  \cdot {x^n}$$
I don't understand why multiplying by $x^{26}$. What's the reason?
 [/INST] You are perfectly right. Using the notation $[x^n] f(x)$ for "the coefficient of $x^n$ in (the series expansion for) $f(x)$" we can write:
\begin{align}
[x^{26}] \left( \frac{1 + x^{10}}{(1 - x)^7} \right)^2
  &= [x^{26}] (1 + x^{10})^2 \cdot (1 - x)^{-14} \\
  &= [x^{26}] (1 +  2 x^{10} + x^{20}) 
                \cdot \sum_{n \ge 0} (-1)^n \binom{-14}{n} x^n \\
  &= [x^{26}] (1 +  2 x^{10} + x^{20}) 
                \cdot \sum_{n \ge 0} \binom{13 + n}{13} x^n \\
  &= ([x^{26}] + 2 [x^{16}] + [x^6]) \sum_{n \ge 0} \binom{13 + n}{13} x^n \\
  &= \binom{39}{13} + 2 \binom{29}{13} + \binom{19}{13}
\end{align}
Looks like a typo of some sort to me (perhaps forgot the $[]$ for "coefficient of" in the above derivation?).
</s>