Đáp án:
$S = {23.2^{20}}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
S=\sum\limits_{k = 0}^{21} {\left( {k + 1} \right)C_{21}^k} \\
= \sum\limits_{k = 0}^{21} {kC_{21}^k} + \sum\limits_{k = 0}^{21} {C_{21}^k} \\
= \sum\limits_{k = 1}^{21} {kC_{21}^k} + \sum\limits_{k = 0}^{21} {C_{21}^k} \\
= \sum\limits_{k = 1}^{21} {21C_{20}^{k - 1}} + {\left( {1 + 1} \right)^{21}}\left( {sd:kC_n^k = nC_{n - 1}^{k - 1}} \right)\\
= 21\sum\limits_{k = 1}^{21} {C_{20}^{k - 1}} + {2^{21}}\\
= 21.{\left( {1 + 1} \right)^{20}} + {2^{21}}\\
= {2^{20}}\left( {21 + 2} \right)\\
= {23.2^{20}}
\end{array}$
