<s>[INST] Prove that $p_2(n) = \left \lfloor{\frac{n}{2}}\right \rfloor+1$ using identity Prove that $p_2(n) = \left \lfloor{\frac{n}{2}}\right \rfloor+1$ using the identity $$\frac{1}{(1-x)(1-x^2)}=\frac{1}{2}\left(\frac{1}{(1-x)^2}\right)+\frac{1}{2}\left(\frac{1}{1-x^2}\right)$$
where $p_k(n)$ is the number of partitions of an integer $n$ into a most $k$ parts. The generating function $P_k(x)$ of $\{p_k(n)\}$ is $$P_k(x) = \frac{1}{\prod_{r=0}^{k}(1-x^k)}$$
Therefore, the generating function $P_2(x)$ of ${p_2(n)}$ is 
$$P_2(x) = \frac{1}{(1-x)(1-x^2)}$$
Now I see here that we can use the identity given above, but I am confused of how to apply it to prove the desired statement. 
 [/INST] Note that
$$\begin{align}\frac{4}{(1-x)(1-x^2)}&=\frac{1}{1+x}+\frac{1}{1-x}+\frac{2}{(1-x)^2}\\
&=\frac{1}{1+x}+\frac{1}{1-x}+\frac{d}{dx}\left(\frac{2}{1-x}\right)\\
&=\sum_{n=0}^{\infty}(-x)^n+\sum_{n=0}^{\infty}x^n
+2\sum_{n=1}^{\infty}nx^{n-1}\\
&=\sum_{n=0}^{\infty}((-1)^n+1+2(n+1))x^n
\end{align}$$
Can you take it from here?
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