Đáp án:
Bạn tham khảo nhé.
Giải thích các bước giải:
\(\begin{array}{l}
e)\,\,\frac{{C_n^k}}{{k + 1}} = \frac{{C_{n + 1}^{k + 1}}}{{n + 1}}\\
\Leftrightarrow \frac{1}{{k + 1}}.\frac{{n!}}{{k!\left( {n - k} \right)!}} = \frac{1}{{n + 1}}.\frac{{\left( {n + 1} \right)!}}{{\left( {k + 1} \right)!\left( {n - k} \right)!}}\\
\Leftrightarrow \frac{1}{{\left( {k + 1} \right)k!}} = \frac{{n + 1}}{{\left( {n + 1} \right)\left( {k + 1} \right)!}}\\
\Leftrightarrow \frac{1}{{\left( {k + 1} \right)!}} = \frac{1}{{\left( {k + 1} \right)!}}\,\,\,\,\left( {luon\,\,\,dung} \right).\\
f)\,\,\,nC_n^k = \left( {k + 1} \right)C_n^{k + 1} + kC_n^k\\
VP = \left( {k + 1} \right)C_n^{k + 1} + kC_n^k\\
= \left( {k + 1} \right).\frac{{n!}}{{\left( {k + 1} \right)!\left( {n - k - 1} \right)!}} + \frac{{k.n!}}{{k!\left( {n - k} \right)!}}\\
= \frac{{n!}}{{k!\left( {n - k - 1} \right)!}} + \frac{{k.n!}}{{k!\left( {n - k} \right)!}}\\
= \frac{{n!}}{{k!}}\left( {\frac{1}{{\left( {n - k - 1} \right)!}} + \frac{k}{{\left( {n - k} \right)!}}} \right)\\
= \frac{{n!}}{{k!}}.\frac{{n - k + k}}{{\left( {n - k} \right)!}}\\
= \frac{{n!.n}}{{k!\left( {n - k} \right)!}}\\
= nC_n^k\\
\Rightarrow dpcm.
\end{array}\)
