(1-1/4)*(1-1/9)*(1-1/16)*.......*(1-1/10000)=?
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\(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{10000}\right)\)
\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\)
\(=\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\)
\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
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Tính:\(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{10000}\right)\)\(=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{100^2}\right)\)\(=\dfrac{3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{99.101}{100^2}\)
\(=\dfrac{2.3.4...99}{2.3.4...100}.\dfrac{3.4.5.6...101}{2.3.4...100}\)
\(=\)\(\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{1}{4}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{9}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{16}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
.............................................................
\(\frac{1}{10000}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)
\(\Rightarrow\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}< 1\)(đpcm)
Xin tk
Ta có: +++.....+=++.......+
mà ++.......+ < ++.......+
++.......+ < 1-+-+.......+-
++.......+ < 1-
++.......+ < 1
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S=1/4+1/9+1/16+...+1/10000 = 1/2x 2 + 1/3x3+...+1/100x100 < 1/1x2 + 1/2x3 +...+ 1/9x10 = 1 - 1/2 + 1/2 - 1/3 +...+ 1/9 - 1/10 = 1- 1/10 < 1
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S=1/4+1/9+1/16+...+1/10000
= 1/2x 2 + 1/3x3+...+1/100x100 < 1/1x2 + 1/2x3 +...+ 1/9x10
= 1 - 1/2 + 1/2 - 1/3 +...+ 1/9 - 1/10 = 1- 1/10 < 1
![](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
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\(x = {6 {} \over 11}\)