Tổng phép tính \[S = C_{10}^6 + C_{10}^7 + C_{10}^8 + C_{10}^9 + C_{10}^{10}.\] bằng

Xét tổng \({\left( {1 + x} \right)^{10}} = C_{10}^0 + xC_{10}^1 + {x^2}C_{10}^2 + ... + {x^{10}}C_{10}^{10}\).

Thay \(x = 1\) ta được: \(C_{10}^0 + C_{10}^1 + C_{10}^2 + ... + C_{10}^{10} = {2^{10}}\).

\(\begin{array}{l} \Rightarrow C_{10}^5 + 2\left( {C_{10}^6 + C_{10}^7 + C_{10}^8 + C_{10}^9 + C_{10}^{10}} \right) = {2^{10}}\\ \Rightarrow C_{10}^6 + C_{10}^7 + C_{10}^8 + C_{10}^9 + C_{10}^{10} = \frac{{{2^{10}} - C_{10}^5}}{2} = 386\end{array}\)