Cho B=8/9 + 24/25 + 48/49 + ... + 200.202/2012. CMR B>99,75.
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13 tháng 6 2017
Ta có:
\(\frac{n\left(n+2\right)}{\left(n+1\right)^2}=1-\frac{1}{\left(n+1\right)^2}>1-\frac{1}{n\left(n+2\right)}=1+\frac{1}{2}.\left(\frac{1}{n+2}-\frac{1}{n}\right)\)
Thế vô bài toán ta được
\(B=\frac{2.4}{3^2}+\frac{4.6}{5^2}+...+\frac{200.202}{201^2}\)
\(>1+1+...+1+\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{2}+\frac{1}{6}-\frac{1}{4}+...+\frac{1}{202}-\frac{1}{200}\right)\)
\(=100+\frac{1}{2}.\left(\frac{1}{202}-\frac{1}{2}\right)=\frac{10075}{101}>99,75\)
TN
13 tháng 6 2017
Ta có đánh giá sau:\(\frac{n\left(n+2\right)}{\left(n+1\right)^2}=1-\frac{1}{\left(n+1\right)^2}\)
\(>1-\frac{1}{x\left(x+2\right)}=1-\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)\)
Suy ra \(B=\frac{2\cdot4}{3^2}+\frac{4\cdot6}{5^2}+\frac{6\cdot8}{7^2}+...+\frac{200\cdot202}{201^2}\)
\(>1-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)+1-\frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}\right)+...+1-\frac{1}{2}\left(\frac{1}{200}-\frac{1}{202}\right)\)
\(=100-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{200}-\frac{1}{202}\right)\)
\(=100-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{202}\right)\)\(=100-\frac{1}{2}\cdot\frac{50}{101}\)
\(>100-\frac{1}{2}\cdot\frac{50}{100}=100-0,25=99,75\)
Tức là \(B>99,75\)
B = \(\frac{8}{9}+\frac{24}{25}+\frac{48}{49}+...+\frac{200.202}{201^2}=\left(1-\frac{1}{3^2}\right)+\left(1-\frac{1}{5^2}\right)+\left(1-\frac{1}{7^2}\right)+...+\left(1-\frac{1}{201^2}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{201^2}\right)\)
\(=100-\left(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{201^2}\right)\)
Ta có Đặt \(C=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+....+\frac{1}{201^2}\)\
\(< \frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{199.201}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{199.201}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{199}-\frac{1}{201}\right)=\frac{1}{2}\left(1-\frac{1}{201}\right)=\frac{1}{2}.\frac{200}{201}=\frac{100}{201}< \frac{1}{2}\)
=> C < 1/2
=> B > 100 - 1/2
=> B > 99,5