Tổng \( C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{1009}\)
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Lời giải:
Báo saiTa có:
\( C_{2019}^0 + C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{2009} = {2^{2019}}\)
Mà
\(\begin{array}{l} C_{2019}^0 = C_{2019}^{2019} = 1,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} C_{2019}^1 = C_{2019}^{2018},{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} C_{2019}^2 = C_{2019}^{2017},{\mkern 1mu} {\mkern 1mu} ...{\mkern 1mu} ,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} C_{2019}^{1009} = C_{2019}^{1010}\\ \begin{array}{*{20}{l}} { \Leftrightarrow 1 + C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{1009} + C_{2019}^{1010} + ... + C_{2019}^1 + 1 = {2^{2019}}}\\ { \Leftrightarrow 2 + 2\left( {C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{1009}} \right) = {2^{2019}}}\\ { \Leftrightarrow C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{1009} = {2^{2018}} - 1.} \end{array} \end{array}\)
