Đáp án:
$C_n^0 + C_n^2 + C_n^4 + ... = C_n^1 + C_n^3 + C_n^5 + ...$
Giải thích các bước giải:
$\begin{array}{l}
Ta\,co:{\left( {1 - x} \right)^n} = \sum\limits_{k = 0}^n {C_n^k{{.1}^{n - k}}.{{\left( { - x} \right)}^k}} = \sum\limits_{k = 0}^n {C_n^k.{{\left( { - 1} \right)}^k}.{x^k}} \\
Cho\,x = 1\,ta\,duoc\,{\left( {1 - 1} \right)^n} = \sum\limits_{k = 0}^n {C_n^k.{{\left( { - 1} \right)}^k}{{.1}^k}} = \sum\limits_{k = 0}^n {C_n^k.{{\left( { - 1} \right)}^k}} \\
\Rightarrow {0^n} = C_n^0 - C_n^1 + C_n^2 - C_n^3 + ... + {\left( { - 1} \right)^n}.C_n^n\\
\Leftrightarrow 0 = C_n^0 - C_n^1 + C_n^2 - C_n^3 + ... + {\left( { - 1} \right)^n}.C_n^n\\
\Leftrightarrow C_n^0 + C_n^2 + C_n^4 + ... = C_n^1 + C_n^3 + C_n^5 + ...
\end{array}$
