Cho M=\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\).Chứng minh rằng: M>18
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A=(1-\(\frac{1}{4}\))+(1-\(\frac{1}{9}\))+(1-\(\frac{1}{16}\))+...+(1-\(\frac{1}{400}\)).
A=19-(\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\))
Ta thấy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}<1\)
=>A>19-1=18(đpcm)
\(a)\)\(M=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)
\(M=\frac{4-1}{4}+\frac{9-1}{9}+\frac{16-1}{16}+...+\frac{400-1}{400}\)
\(M=1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+1-\frac{1}{400}\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{400}\right)\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)\)
Do từ 2 đến 20 có \(20-2+1=19\) nên :
\(M=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\)
\(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)
\(A>\frac{1}{2}-\frac{1}{21}\)
\(\Rightarrow\)\(M=19-A>19-\frac{1}{2}+\frac{1}{21}=18,5+\frac{1}{21}>8\)
\(\Rightarrow\)\(M>8\) ( đpcm )
Còn câu b) bn xem lại đề đi, nếu đề đúng thì mk sai :v
Chúc bạn học tốt ~
\(M=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}...+\frac{399}{400}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+\left(1-\frac{1}{25}\right)+...+\left(1-\frac{1}{400}\right)\)
\(=\left(1+1+1+....+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{20^2}\right)\)
\(=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)\)
Đặt \(N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{20^2}\)
\(< P=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{20\cdot21}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{20}-\frac{1}{21}\)
\(=\frac{1}{2}-\frac{1}{21}\)
\(\Rightarrow M+N>19-\frac{1}{2}+\frac{1}{21}=\frac{37}{2}+\frac{1}{21}>8\)
b sai đề.chừng nào chữa đề thì làm
Xét A= \(\frac{3}{4}\)+ \(\frac{8}{9}\) +...+ \(\frac{399}{400}\)
= (1 - \(\frac{1}{2^2}\)) + (1- \(\frac{1}{3^2}\)) +...+ (1- \(\frac{1}{20^2}\))
= (1+1+1+...+1) - (\(\frac{1}{2^2}\) + \(\frac{1}{3^2}\)+...+ \(\frac{1}{20^2}\)) Bạn phải mở ngoặc có 19 số 1 nha!
= 19 - (\(\frac{1}{2^2}\) + \(\frac{1}{3^2}\)+...+ \(\frac{1}{20^2}\))
Đặt B =\(\frac{1}{2^2}\) + \(\frac{1}{3^2}\)+...+ \(\frac{1}{20^2}\) < \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +...+ \(\frac{1}{19.20}\) = 1- \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) +...+ \(\frac{1}{19}\) - \(\frac{1}{20}\) = 1 - \(\frac{1}{20}\) = \(\frac{19}{20}\)
=> A= 19 - B= 18+ 1- \(\frac{19}{20}\) >18 => A>18
\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{400}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{400}\right)\)
\(=20-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)< 20\) (1)
Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
.......
\(\frac{1}{20^2}< \frac{1}{19.20}=\frac{1}{19}-\frac{1}{20}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)
\(\Rightarrow A>20-1=19\) (2)
Từ (1) và (2) => 19 < A < 20
Vậy...
Ta có : \(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{15}+...+\frac{1}{10000}\right)\)
\(=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)< 99\)
\(\Rightarrow\)S<99 (1)
Đặt \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\)
\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)
...
\(\frac{1}{100^2}=\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}< 1\)
\(\Rightarrow\)S>99-1=98 (2)
Từ (1) và (2)
\(\Rightarrow\)98<S<99
\(\Rightarrow\)S\(\notin\)N
Vậy S\(\notin\)N.
\(M=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)
\(\Rightarrow M=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+....+\frac{20^2-1}{20^2}\)
\(\Rightarrow M=\frac{2^2}{2^2}-\frac{1}{2^2}+\frac{3^2}{3^2}-\frac{1}{3^2}+\frac{4^2}{4^2}-\frac{1}{4^2}+....+\frac{20^2}{20^2}-\frac{1}{20^2}\)
\(\Rightarrow M=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{20^2}\right)\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{20^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{19\cdot20}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{19}{20}< 1\)
\(\Rightarrow A< 1\)
\(\Rightarrow M>18\)