Let $\alpha(n)$ be the number of ways to write a positive integer $n$ as the sum of $1$s and $2$s. Let $\beta(n)$ be the number of ways to write $n$ as a sum of several integers greater than $1$. Different orders are treated as different. Prove $\alpha(n)=\beta(n+2)$.
If $\alpha(n)$ is the number of ways which $n$ can be expressed as the sum of $k$ terms of either $1$ or $2$, then $\alpha(n)$ must be the coefficient of the term in the expanded form of $x^n$ in $(x+x^2)^k$. This means that $$1+\sum_{n=1}^{\infty}\alpha(n)x^n = \sum_{k=0}^{\infty}(x+x^2)^k=\frac{1}{1-x-x^2}$$
Similarly, we have $$\begin{array}{rl} \displaystyle 1+\sum_{n=2}^{\infty}\beta(n)x^n=&\displaystyle\sum_{k=0}^{\infty}(x^2+x^3+\cdots)^k \\= & \displaystyle\sum_{k=0}^{\infty}\left(\frac{x^2}{1-x}\right)^k \\ =&\displaystyle \left(1-\frac{x^2}{1-x}\right)^{-1}\\=& \displaystyle \frac{1-x}{1-x-x^2} \\ =& \displaystyle 1 + \frac{x^2}{1-x-x^2} \end{array}$$
Therefore, we have $$\sum_{n=2}^{\infty}\beta(n)x^n = x^2 + x^2\sum_{n=1}^{\infty}\alpha(n)x^n$$
Comparing the coefficients of $x^n$ immediately gives the desired result.