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\(A=1\frac{4}{5}\times1\frac{4}{21}\times1\frac{4}{45}\times...\)
Ta có: \(1\frac{4}{5}=\frac{5+4}{5}=\frac{9}{5}=\frac{3\times3}{1\times5}\)
\(1\frac{4}{21}=\frac{21+4}{21}=\frac{25}{21}=\frac{5\times5}{3\times7}\)
\(1\frac{4}{45}=\frac{45+4}{45}=\frac{49}{45}=\frac{7\times7}{5\times9}\)
...
Tổng quát thừa số thứ \(n\)là: \(\frac{\left(2\times n+1\right)\times\left(2\times n+1\right)}{\left(2\times n-1\right)\times\left(2\times n+3\right)}\)
Thừa số thứ \(100\)là: \(\frac{201\times201}{199\times203}\).
Tích \(A\)là:
\(A=\frac{3\times3}{1\times5}\times\frac{5\times5}{3\times7}\times\frac{7\times7}{5\times9}\times...\times\frac{201\times201}{199\times203}\)
\(=\frac{\left(3\times5\times7\times...\times201\right)\times\left(3\times5\times7\times...\times201\right)}{\left(1\times3\times5\times...\times199\right)\times\left(5\times7\times9\times...\times203\right)}\)
\(=\frac{201\times3}{1\times203}=\frac{603}{203}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
trả lời
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cám ơn các bn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(N=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}\)
\(N>\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}\)
\(N>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{10}{22}>\frac{9}{22}\)
Vậy N > 9/22
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1/4 +1/9 + 1/16 + 1/25 + 1/36
= ( 1/4 + 1/16 ) + ( 1/9 + 1/36) + 1/25
= 5/16 + 5/36 + 1/25
= 65/144 + 1/25
= 1769/3600
=> 1769/3600 < 5/6 (hay 1769/3600 < 3000/3600 -quyđồng-)
Vậy A< 5/6
Đúng nhé, tk cho mjk với-số to thiệt nhưng đúng mà-
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1/(2x2)+1/(3x3)+...+1/(100x100)
Nhận thấy rằng n x n -1=n x n -n+n-1=n x (n-1)+n-1=(n-1) x (n+1)
=> A < 1/(2x2-1)+1/(3x3-1)+...+1/(100x100-1)=1/(1x3)+1/(3x5)+...+1/(99x101)=1/2-1/202<1/2<3/4
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1/(2x2)+1/(3x3)+...+1/(100x100) Nhận thấy rằng n x n -1=n x n -n+n-1=n x (n-1)+n-1=(n-1) x (n+1) => A < 1/(2x2-1)+1/(3x3-1)+...+1/(100x100-1)=1/(1x3)+1/(3x5)+...+1/(99x101)=1/2-1/202<1/2<3/4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{4036081}\)
\(=\frac{1}{2\times2}+\frac{1}{3\times3}+\frac{1}{4\times4}+...+\frac{1}{2009\times2009}\)
\(< \frac{1}{2\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2008\times2009}\)
\(=\frac{1}{4}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+...+\frac{2009-2008}{2008\times2009}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\)
\(=\frac{3}{4}-\frac{1}{2009}< \frac{3}{4}\)