Chủ đề này có 1 trả lời

[the man]

Tính tổng:

1. $A=\textrm{C}_{2009}^{0}-\frac{1}{3}\textrm{C}_{2009}^{2}+\frac{1}{5}\textrm{C}_{2009}^{4}-...+\frac{1}{2009}\textrm{C}_{2009}^{2008}$

2.$B=\frac{\textrm{C}_{8}^{8}}{7.8}+\frac{\textrm{C}_{9}^{8}}{8.9}+...+\frac{\textrm{C}_{2010}^{8}}{2009.2010}$

[ChiLanA0K48]

Tính tổng:

1. $A=\textrm{C}_{2009}^{0}-\frac{1}{3}\textrm{C}_{2009}^{2}+\frac{1}{5}\textrm{C}_{2009}^{4}-...+\frac{1}{2009}\textrm{C}_{2009}^{2008}$

2.$B=\frac{\textrm{C}_{8}^{8}}{7.8}+\frac{\textrm{C}_{9}^{8}}{8.9}+...+\frac{\textrm{C}_{2010}^{8}}{2009.2010}$

 

Câu 1:

$A=\sum_{k=0} ^{1004}\frac{(-1)^k}{2k+1}C_{2009}^{2k}=\frac{1}{2010}\sum_{k=0}^{1004}C_{2010}^{2k+1}.(-1)^k$

Xét 

$(1+i)^{2010}=\sum_{k=0}^{2010}C_{2010}^{k}i^{k}=\sum_{j=0}^{1005}C_{2010}^{2j}i^{2j}+\sum_{k=0}^{1004}C_{2010}^{2k+1}i^{2k+1}=\sum_{j=0}^{1005}C_{2010}^{2j}.(-1)^{j}+(\sum_{k=0}^{1004}C_{2010}^{2k+1}(-1)^{k})i$

Mặt khác, cũng có

$(1+i)^{2010}=\left [ \sqrt{2}\left ( cos\frac{\pi}{4}+i.sin\frac{\pi}{4} \right ) \right ]^{2010}=2^{1005}cos\frac{2010\pi}{4}+\left ( 2^{1005}sin\frac{2010\pi}{4} \right )i=2^{1005}i$

Từ trên suy ra 

$\sum_{k=0}^{1004}C_{2010}^{2k+1}.(-1)^k=2^{1005}$

Vậy $A=\frac{1}{2010}\sum_{k=0}^{1004}C_{2010}^{2k+1}.(-1)^k=\frac{2^{1005}}{2010}$

 

Câu 2:

$B=\sum_{k=8}^{2010}\frac{C_{k}^{8}}{(k-1)k}=\frac{1}{8}\sum_{k=8}^{2010}\frac{C_{k-1}^{7}}{k-1}=\frac{1}{56}\sum_{k=8}^{2010}C_{k-2}^{6}=\frac{1}{56}sum_{i=6}^{2008}C_{i}^{6}$

 

Xét $S=\sum_{k=6}^{2008}(x+1)^k=\frac{(x+1)^{2009}-(x+1)^6}{x}=$

Hệ số của $x^6$ trong khai triển $S$ là  $\sum_{i=6}^{2008}C_{i}^{6}=C_{2009}^{7}$

Vậy $B=\frac{C_{2009}^{7}}{56}$