\displaystyle\sum^n_{m=0}\left({\begin{array}{c}n\\ m\\[br] \end{array}} \right) = 2^n
\displaystyle\sum^n_{m=0}\left({\begin{array}{c}n\\ m\\[br] \end{array}} \right) = \left({\begin{array}{c}n\\ 0\\[br] \end{array}} \right) + \left({\begin{array}{c}n\\ 1\\[br] \end{array}} \right) + ... + \left({\begin{array}{c}n\\ n-1\\[br] \end{array}} \right) + \left({\begin{array}{c}n\\ n\\[br] \end{array}} \right)
= 1 + \left({\begin{array}{c}n\\ 1\\[br] \end{array}} \right) + ... + \left({\begin{array}{c}n\\ n-1\\[br] \end{array}} \right) + 1 = 2 + 2 \left( \left({\begin{array}{c}n\\ 1\\[br] \end{array}} \right) + ... \right)
\displaystyle\sum^n_{m=0}\left({\begin{array}{c}n\\ m\\[br] \end{array}} \right) = 2^n
\displaystyle\sum^n_{m=0}\left({\begin{array}{c}n\\ m\\[br] \end{array}} \right) = \left({\begin{array}{c}n\\ 0\\[br] \end{array}} \right) + \left({\begin{array}{c}n\\ 1\\[br] \end{array}} \right) + ... + \left({\begin{array}{c}n\\ n-1\\[br] \end{array}} \right) + \left({\begin{array}{c}n\\ n\\[br] \end{array}} \right)
= 1 + \left({\begin{array}{c}n\\ 1\\[br] \end{array}} \right) + ... + \left({\begin{array}{c}n\\ n-1\\[br] \end{array}} \right) + 1 = 2 + 2 \left( \left({\begin{array}{c}n\\ 1\\[br] \end{array}} \right) + ... \right)