S= [tex](1/6)c^1_{2004}+(1/4)c^2_{2004}+(3/10)c^3_{2004}+...+(1002/2006)c^{2004}_{2004}[/tex]
kiến thức 12 cùng hơi quên rồi nếu các bạn thấy dễ quá thi các bạn nói để mình nghĩ bai khác cho nghe
[TEX]S=\frac{1}{2}\(\frac{1}{1+2}.C_{2004}^1+\frac{2}{2+2}C_{2004}^2+\frac{3}{3+2}C_{2004}^2+.....+\frac{2004}{2004+2}C_{2004}^{2004}\) [/TEX]
[TEX]\rightarrow2S= \sum_{k=1}^{2004}\frac{k}{k+2}.C_{2004}^k=\sum_{k=1}^{2004}C_{2004}^k- 2\sum_{k=1}^{2004}\frac{1}{k+2}.C_{2004}^k[/TEX]
[TEX]=\(2^{2004}-1\)-2\sum_{k=1}^{2004}\frac{1}{k+2}.C_{2004}^k[/TEX]
Ta luôn có :
[TEX]x(1+x)^{2004} =\sum_{k=0}^{2004} C_{2004}^k x^{k+1}[/TEX]
[TEX]\rightarrow 2\int_{0}^{1}x(1+x)^{2004}dx=2\(\sum_{k=0}^{2004} \frac{1}{k+2} C_{2004}^k x^{k+2}\)_{0}^1=C_{2004}^0+ 2\sum_{k=1}^{2004}\frac{1}{k+2}.C_{2004}^k\ \ (**)[/TEX]
Xét : [TEX]I= 2\int_{0}^{1}x(1+x)^{2004}dx[/TEX]
Đặt: [TEX]u=x+1\righ du=dx[/TEX]
[TEX]I=2\int_{1}^{2}(u-1)u^{2004}du =2\int_{1}^{2} \(u^{2005}-u^{2004}\)du =2\(\frac{2^{2006}}{2006}-\frac{1}{2006}- \frac{2^{2005}}{2005}+\frac{1}{2005}\)= \frac{2^{2007}}{2006}- \frac{2^{2006}}{2005}+\frac{2}{2006.2005}\ \(***)[/TEX]
[TEX](**)&(**)\righ 2\sum_{k=1}^{2004}\frac{1}{k+2}.C_{2004}^k= \frac{2^{2007}}{2006}- \frac{2^{2006}}{2005}+\frac{2}{2006.2005}-1[/TEX]
[TEX](***)&(**)&(*)\rightarrow 2S=\(2^{2004}-1\)-\(\frac{2^{2007}}{2006}- \frac{2^{2006}}{2005}+\frac{2}{2006.2005}-1\)[/TEX]
[TEX]\righ S=2^{2003}-\frac{2^{2006}}{2006}+ \frac{2^{2005}}{2005}-\frac{1}{2006.2005}[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
