Chủ đề này có 1 trả lời

[Mary Huynh]

Tính tổng : 
 $S= \frac{C_{8}^{8}\textrm{}}{7.8} +\frac{C_{9}^{8}\textrm{}}{8. 9}+\frac{C_{10}^{8}\textrm{}}{9.10}+...+\frac{C_{2012}^{8}\textrm{}}{2011.2012}$
 

[chanhquocnghiem]

Tính tổng : 
 $S= \frac{C_{8}^{8}\textrm{}}{7.8} +\frac{C_{9}^{8}\textrm{}}{8. 9}+\frac{C_{10}^{8}\textrm{}}{9.10}+...+\frac{C_{2012}^{8}\textrm{}}{2011.2012}$
 

Ta có :

$8!.S=\frac{A_8^8}{7.8}+\frac{A_9^8}{8.9}+\frac{A_{10}^8}{9.10}+...+\frac{A_{2012}^8}{2011.2012}=A_6^6+A_7^6+A_8^6+...+A_{2010}^6$

Mà $\sum_{k=p}^{q}A_k^p=A_p^p+A_{p+1}^p+A_{p+2}^p+...+A_q^p=\frac{A_{q+1}^{p+1}}{p+1}$ (*)

Do đó $8!.S=A_6^6+A_7^6+A_8^6+...+A_{2010}^6=\frac{A_{2011}^7}{7}$

$\Rightarrow S=\frac{A_{2011}^7}{7.8!}=\frac{C_{2011}^7}{56}$.

 

-----------------------------------------------------

Nếu bạn không thừa nhận công thức (*) thì có thể chứng minh nó như sau :

Đặt vế trái của (*) là $T$, ta có :

$(p+1)T=(p+1)A_p^p+(p+1)A_{p+1}^p+(p+1)A_{p+2}^p+...+(p+1)A_q^p$

$(p+1)A_p^p=1.2.3...(p+1)$

$(p+1)A_{p+1}^p=(p+1)\left [ 2.3.4...(p+1) \right ]=(p+2)\left [ 2.3.4...(p+1) \right ]-1.\left [ 2.3.4...(p+1) \right ]=2.3.4...(p+2)-1.2.3...(p+1)$

$(p+1)A_{p+2}^p=(p+1)\left [ 3.4.5...(p+2) \right ]=(p+3)\left [ 3.4.5...(p+2) \right ]-2.\left [ 3.4.5...(p+2) \right ]=3.4.5...(p+3)-2.3.4...(p+2)$

$(p+1)A_{p+3}^p=(p+1)\left [ 4.5.6...(p+3) \right ]=(p+4)\left [ 4.5.6...(p+3) \right ]-3.\left [ 4.5.6...(p+3) \right ]=4.5.6...(p+4)-3.4.5...(p+3)$

...............................................................

$(p+1)A_q^p=(p+1)\left [ (q-p+1)(q-p+2)...q \right ]=(q+1)\left [ (q-p+1)(q-p+2)...q \right ]-(q-p).\left [ (q-p+1)(q-p+2)...q \right ]=(q-p+1)(q-p+2)...(q+1)-(q-p)(q-p+1)...q$

Cộng tất cả lại, vế theo vế $\Rightarrow (p+1)T=(q-p+1)(q-p+2)...(q+1)=A_{q+1}^{p+1}\Rightarrow T=\frac{A_{q+1}^{p+1}}{p+1}$