Đáp án:
\(S = {2^{19}} - 1\).
Giải thích các bước giải:
Xét khai triển
\(\begin{array}{l}{\left( {x + 1} \right)^{19}} = \sum\limits_{k = 0}^{19} {C_{19}^k{x^k}} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = C_{19}^0 + C_{19}^1x + C_{19}^2{x^2} + ... + C_{19}^{19}{x^{19}}\end{array}\)
Thay \(x = 1\) ta có:
\(\begin{align}{{2}^{19}}=C_{19}^{0}+C_{19}^{1}+C_{19}^{2}+...+C_{19}^{18}+C_{19}^{19} \\ {{2}^{19}}=~C_{19}^{0}+C_{19}^{1}+C_{19}^{2}+...+C_{19}^{18}+1 \\ \Rightarrow C_{19}^{0}+C_{19}^{1}+C_{19}^{2}+...+C_{19}^{18}={{2}^{19}}-1 \\ \end{align}\)
Vậy \(S = {2^{19}} - 1\).
