Chứng minh rằng với mọi $n\in N$, ta luôn có: \[\dfrac{C^0_n}{C^1_{n+2}}+\dfrac{C^1_n}{C^2_{n+3}}+...+\dfrac{C^n_n}{C^{n+1}_{2n+2}}=\dfrac{1}{2}\] Lời giải \[\begin{aligned}\dfrac{C_n^k}{C_{n+2+k}^{k+1}}&=\dfrac{n!}{\left( n-k \right)!k!}.\dfrac{\left( k+1 \right)!\left( n+1 \right)!}{\left( n+2+k \right)!}\\&=\dfrac{k+1}{n+2}.\dfrac{n!\left( n+2\right)!}{\left( n-k \right)!\left( n+2+k\right)!}\\&=\dfrac{1}{2}\left[ \left( n+k+2 \right)-(n-k) \right]\dfrac{\left( 2n+2 \right)!}{\left( n+k+2 \right)!\left( n-k \right)!}.\dfrac{1}{\left( n+2 \right)C_{2n+2}^{n}}\\&=\frac{1}{2}\left[ \frac{\left( 2n+2 \right)!}{\left( n+k+1 \right)!\left( n-k \right)!}-\frac{\left( 2n+2 \right)!}{\left( n+k+2 \right)!\left( n-k-1 \right)!} \right].\frac{1}{\left( n+2 \right)C_{2n+2}^{n}}\\&=\frac{\left( n+1 \right)\left( C_{2n+1}^{n-k}-C_{2n+1}^{n-k-1} \right)}{\left( n+2 \right)C_{2n+2}^{n}}\end{aligned}\] Thay $k=0,1,..,n-1$, ta có: \[\dfrac{C_{n}^{0}}{C_{n+2}^{1}}=\dfrac{\left( n+1 \right)\left( C_{2n+1}^{n}-C_{2n+1}^{n-1} \right)}{\left( n+2\right)C_{2n+2}^{n}}\\\dfrac{C_{n}^{1}}{C_{n+3}^{2}}=\dfrac{\left( n+1 \right)\left( C_{2n+1}^{n-1}-C_{2n+1}^{n-2} \right)}{\left( n+2 \right)C_{2n+2}^{n}}\\....................................................\\\dfrac{C_{n}^{n-1}}{C_{2n+2}^{n}}=\dfrac{\left( n+1 \right)\left( C_{2n+1}^{1}-C_{2n+1}^{0} \right)}{\left( n+2 \right)C_{2n+2}^{n}}\] \[\begin{aligned}\Rightarrow \frac{C_{n}^{0}}{C_{n+2}^{1}}+\frac{C_{n}^{1}}{C_{n+3}^{2}}+...+\frac{C_{n}^{n-1}}{C_{2n+2}^{n}}+\frac{C_{n}^{n}}{C_{2n+2}^{n+1}}&=\frac{\left( n+1 \right)\left( C_{2n+1}^{n}-C_{2n+1}^{0} \right)}{\left( n+2 \right)C_{2n+2}^{n}}+\frac{C_{n}^{n}}{C_{2n+2}^{n+1}}\\&=\frac{\dfrac{1}{2}\left( 2n+2 \right)\left[ \dfrac{\left( 2n+1 \right)!}{n!\left( n+2 \right)!}-\dfrac{1}{n+2} \right]}{C_{2n+2}^{n}}+\frac{\left( n+1 \right)!\left( n+1 \right)!}{\left( 2n+2 \right)!}\\&=\frac{\dfrac{1}{2}\dfrac{\left( 2n+2 \right)!}{n!\left( n+2 \right)!}}{C_{2n+2}^{n}}-\frac{n+1}{n+2}\frac{n!\left( n+2 \right)!}{\left( 2n+2 \right)!}+\frac{\left( n+1 \right)!\left( n+1 \right)!}{\left( 2n+2 \right)!}\\&=\frac{1}{2}-\frac{\left( n+1 \right)!\left( n+1 \right)!}{\left( 2n+2 \right)!}+\frac{\left( n+1 \right)!\left( n+1 \right)!}{\left( 2n+2 \right)!}\\&=\frac{1}{2}\end{aligned}\]

Combination 02

Chứng minh rằng với mọi $n \in N$ , ta luôn có:

$\frac{C _{n}^{0}}{C _{n + 2}^{1}} + \frac{C _{n}^{1}}{C _{n + 3}^{2}} + ... + \frac{C _{n}^{n}}{C _{2 n + 2}^{n + 1}} = \frac{1}{2}$

Lời giải

$\frac{C _{n}^{k}}{C _{n + 2 + k}^{k + 1}} = \frac{n !}{( n - k ) ! k !} . \frac{( k + 1 ) ! ( n + 1 ) !}{( n + 2 + k ) !} = \frac{k + 1}{n + 2} . \frac{n ! ( n + 2 ) !}{( n - k ) ! ( n + 2 + k ) !} = \frac{1}{2} [(n + k + 2) - (n - k)] \frac{( 2 n + 2 ) !}{( n + k + 2 ) ! ( n - k ) !} . \frac{1}{( n + 2 ) C _{2 n + 2}^{n}} = \frac{1}{2} [\frac{( 2 n + 2 ) !}{( n + k + 1 ) ! ( n - k ) !} - \frac{( 2 n + 2 ) !}{( n + k + 2 ) ! ( n - k - 1 ) !}] . \frac{1}{( n + 2 ) C _{2 n + 2}^{n}} = \frac{( n + 1 ) ( C _{2 n + 1}^{n - k} - C _{2 n + 1}^{n - k - 1} )}{( n + 2 ) C _{2 n + 2}^{n}}$

Thay $k = 0, 1, .., n - 1$ , ta có:

$\frac{C _{n}^{0}}{C _{n + 2}^{1}} = \frac{( n + 1 ) ( C _{2 n + 1}^{n} - C _{2 n + 1}^{n - 1} )}{( n + 2 ) C _{2 n + 2}^{n}} \frac{C _{n}^{1}}{C _{n + 3}^{2}} = \frac{( n + 1 ) ( C _{2 n + 1}^{n - 1} - C _{2 n + 1}^{n - 2} )}{( n + 2 ) C _{2 n + 2}^{n}} .................................................... \frac{C _{n}^{n - 1}}{C _{2 n + 2}^{n}} = \frac{( n + 1 ) ( C _{2 n + 1}^{1} - C _{2 n + 1}^{0} )}{( n + 2 ) C _{2 n + 2}^{n}}$

$\Rightarrow \frac{C _{n}^{0}}{C _{n + 2}^{1}} + \frac{C _{n}^{1}}{C _{n + 3}^{2}} + ... + \frac{C _{n}^{n - 1}}{C _{2 n + 2}^{n}} + \frac{C _{n}^{n}}{C _{2 n + 2}^{n + 1}} = \frac{( n + 1 ) ( C _{2 n + 1}^{n} - C _{2 n + 1}^{0} )}{( n + 2 ) C _{2 n + 2}^{n}} + \frac{C _{n}^{n}}{C _{2 n + 2}^{n + 1}} = \frac{\frac{1}{2} ( 2 n + 2 ) [ \frac{( 2 n + 1 ) !}{n ! ( n + 2 ) !} - \frac{1}{n + 2} ]}{C _{2 n + 2}^{n}} + \frac{( n + 1 ) ! ( n + 1 ) !}{( 2 n + 2 ) !} = \frac{\frac{1}{2} \frac{( 2 n + 2 ) !}{n ! ( n + 2 ) !}}{C _{2 n + 2}^{n}} - \frac{n + 1}{n + 2} \frac{n ! ( n + 2 ) !}{( 2 n + 2 ) !} + \frac{( n + 1 ) ! ( n + 1 ) !}{( 2 n + 2 ) !} = \frac{1}{2} - \frac{( n + 1 ) ! ( n + 1 ) !}{( 2 n + 2 ) !} + \frac{( n + 1 ) ! ( n + 1 ) !}{( 2 n + 2 ) !} = \frac{1}{2}$

Combination 02

Đề bài

Mô tả