By considering the expansion of $\left(1+x\right)^{n}$ where $n$ is a positive integer, or otherwise, show that:

• $ \displaystyle \begin{pmatrix}n\\0\end{pmatrix} + \begin{pmatrix}n\\1\end{pmatrix} + \begin{pmatrix}n\\2\end{pmatrix} + \cdots + \begin{pmatrix}n\\n\end{pmatrix}=2^{n} ; $
• $ \displaystyle \begin{pmatrix}n\\1\end{pmatrix} + 2\begin{pmatrix}n\\2\end{pmatrix} + 3\begin{pmatrix}n\\3\end{pmatrix} +\cdots + n\begin{pmatrix}n\\n\end{pmatrix}=n2^{n-1} ; $
• $ \displaystyle \begin{pmatrix}n\\0\end{pmatrix} + \dfrac{1}{2}\begin{pmatrix}n\\1\end{pmatrix} + \dfrac{1}{3}\begin{pmatrix}n\\2\end{pmatrix} + \cdots + \dfrac{1}{n+1}\begin{pmatrix}n\\n\end{pmatrix} = \dfrac{1}{n+1}\left(2^{n+1}-1\right); $
• $ \displaystyle \begin{pmatrix}n\\1\end{pmatrix} + 2^{2}\begin{pmatrix}n\\2\end{pmatrix} + 3^{2}\begin{pmatrix}n\\3\end{pmatrix} +\cdots + n^{2}\begin{pmatrix}n\\n\end{pmatrix} = n\left(n+1\right)2^{n-2}. $