Biểu thức \(2C_n^k + 5C_n^{k + 1} + 4C_n^{k + 2}+C_n^{k+3}\) bằng biểu thức nào sau đây?

Trước hết ta chứng minh \(C_n^k + C_n^{k + 1} = C_{n + 1}^{k + 1}\)

\(\begin{array}{l}VT = C_n^k + C_n^{k + 1}\\ = \dfrac{{n!}}{{k!\left( {n - k} \right)!}} + \dfrac{{n!}}{{\left( {k + 1} \right)!\left( {n - k - 1} \right)!}}\\ = \dfrac{{n!}}{{k!\left( {n - k - 1} \right)!}}\left( {\dfrac{1}{{n - k}} + \dfrac{1}{{k + 1}}} \right)\\ = \dfrac{{n!}}{{k!\left( {n - k - 1} \right)!}}.\dfrac{{k + 1 + n - k}}{{\left( {n - k} \right)\left( {k + 1} \right)}}\\ = \dfrac{{n!\left( {n + 1} \right)}}{{k!\left( {k + 1} \right)\left( {n - k - 1} \right)!\left( {n - k} \right)}}\\ = \dfrac{{\left( {n + 1} \right)!}}{{\left( {k + 1} \right)!\left( {n - k} \right)!}} = C_{n + 1}^{k + 1} = VP\end{array}\)

Ta có:

\(\begin{array}{l}\,\,\,\,C_n^k + 2C_n^{k + 1} + C_n^{k + 2}\\ = C_n^k + C_n^{k + 1} + C_n^{k + 1} + C_n^{k + 2}\\ = C_{n + 1}^{k + 1} + C_{n + 1}^{k + 2}\\ = C_{n + 2}^{k + 2}\\\,\,\,\,C_n^k + 3C_n^{k + 1} + 3C_n^{k + 2} + C_n^{k + 3}\\ = C_n^k + C_n^{k + 1} + 2\left( {C_n^{k + 1} + C_n^{k + 2}} \right) + C_n^{k + 2} + C_n^{k + 3}\\ = C_{n + 1}^{k + 1} + 2C_{n + 1}^{k + 2} + C_{n + 1}^{k + 3}\\ = C_{n + 1}^{k + 1} + C_{n + 1}^{k + 2} + C_{n + 1}^{k + 2} + C_{n + 1}^{k + 3}\\ = C_{n + 2}^{k + 2} + C_{n + 2}^{k + 3}\\ = C_{n + 3}^{k + 3}\\ \Rightarrow 2C_n^k + 5C_n^{k + 1} + 4C_n^{k + 2} + C_n^{k + 3}= C_{n + 2}^{k + 2} + C_{n + 3}^{k + 3}\end{array}\)