Chủ đề này có 2 trả lời

[thuytop]

Chứng minh rằng :
$C = \left ( 1 + \frac{4}{5} \right ) \left ( 1 + \frac{4}{12} \right ) \left ( 1 + \frac{4}{21} \right )...\left ( 1 + \frac{4}{n(n+4)} \right ) < 6$ với $n \in \mathbb{N^{*}}$

Bài viết đã được chỉnh sửa nội dung bởi tramyvodoi: 16-12-2012 - 11:27

[DarkBlood]

Chứng minh rằng :
$C = \left ( 1 + \frac{4}{5} \right ) \left ( 1 + \frac{4}{12} \right ) \left ( 1 + \frac{4}{21} \right )...\left ( 1 + \frac{4}{n(n+4)} \right ) < 6$ với $n \in \mathbb{N^{*}}$

Ta có:
$C = \left ( 1 + \frac{4}{5} \right ) \left ( 1 + \frac{4}{12} \right ) \left ( 1 + \frac{4}{21} \right )...\left ( 1 + \frac{4}{n(n+4)} \right )$

$=\left ( \frac{3^2}{1.5} \right )\left ( \frac{4^2}{2.6} \right )\left ( \frac{5^2}{3.7} \right )\left ( \frac{6^2}{4.8} \right )...\left [ \frac{\left ( n+2 \right )^2}{n\left ( n+4 \right )} \right ]$
Tách $C=A.B$
Trong đó:
$A=\left ( \frac{3^2}{1.5} \right )\left ( \frac{5^2}{3.7} \right )...\left [ \frac{\left ( n+1 \right )^2}{(n-1)(n+3)} \right ]$

$B=\left ( \frac{4^2}{2.6} \right )\left ( \frac{6^2}{4.8} \right )...\left [ \frac{\left ( n+2 \right )^2}{n(n+4)} \right ]$

Ta có:
$A=\left ( \frac{3^2}{1.5} \right )\left ( \frac{5^2}{3.7} \right )...\left [ \frac{\left ( n+1 \right )^2}{(n-1)(n+3)} \right ]$

$=\left ( \frac{3}{1}.\frac{5}{3}.\frac{7}{5}...\frac{n+1}{n-1} \right )\left ( \frac{3}{5}.\frac{5}{7}.\frac{7}{9}...\frac{n+1}{n+3} \right )$

$=\left ( n+1 \right ).\frac{3}{n+3}=\frac{3n+3}{n+3}<\frac{3n+9}{n+3}=3$

$B=\left ( \frac{4^2}{2.6} \right )\left ( \frac{6^2}{4.8} \right )...\left [ \frac{\left ( n+2 \right )^2}{n(n+4)} \right ]$

$=\left ( \frac{4}{2}.\frac{6}{4}.\frac{8}{6}...\frac{n+2}{n} \right )\left ( \frac{4}{6}.\frac{6}{8}.\frac{8}{10}...\frac{n+2}{n+4} \right )$

$=\frac{n+2}{2}.\frac{4}{n+4}=2.\frac{n+2}{n+4}<2.\frac{n+4}{n+4}=2$

Do đó: $C=A.B<2.3=6$

Bài viết đã được chỉnh sửa nội dung bởi Hoang Huy Thong: 16-12-2012 - 12:47

[DarkBlood]

Chứng minh rằng :
$C = \left ( 1 + \frac{4}{5} \right ) \left ( 1 + \frac{4}{12} \right ) \left ( 1 + \frac{4}{21} \right )...\left ( 1 + \frac{4}{n(n+4)} \right ) < 6$ với $n \in \mathbb{N^{*}}$

Bài này còn có thể chứng mình $C>1.$
Ta có:
$C=\left ( \frac{3^2}{1.5} \right )\left ( \frac{4^2}{2.6} \right )\left ( \frac{5^2}{3.7} \right )\left ( \frac{6^2}{4.8} \right )...\left [ \frac{\left ( n+2 \right )^2}{n\left ( n+4 \right )} \right ]$
Nhận xét:
$k(k+4)=k^2+4k<k^2+4k+3=(k+1)(k+3)$
Do đó:
$C>\left ( \frac{3^2}{2.4} \right )\left ( \frac{4^2}{3.5} \right )\left ( \frac{5^2}{4.6} \right )\left ( \frac{6^2}{5.7} \right )...\left [ \frac{\left ( n+2 \right )^2}{(n+1)\left ( n+3 \right )} \right ]$

$=\left ( \frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}...\frac{n+2}{n+1} \right )\left ( \frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{n+2}{n+3} \right )$

$=\frac{n+2}{2}.\frac{3}{n+3}=\frac{3n+6}{2n+6}>\frac{3n+6}{3n+6}=1$
Vậy $C>1$

Bài viết đã được chỉnh sửa nội dung bởi Hoang Huy Thong: 16-12-2012 - 13:34