BT
1
K
Khách
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NP
3
24 tháng 5 2017
đề đúng rồi
\(C=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(C=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(C=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
\(C=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
đặt \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow A< 1\)
Vì \(A< 1\)nên \(B=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)>99-1=98\)
24 tháng 5 2017
= 3/22 + 8/32 + 15/42 + ... + 9999/1002
= 1.3/2.2 + 2.4/3.3 + 3.5/4.4 + .... + 99.101/100.100
\(=\frac{1.3.2.4.3.5.4.6...99.101}{2^2.3^2....100^2}=\frac{1.2.3^2.4^2...99^2.100.101}{2^2.3^2....100^2}=\frac{1.2.101}{2^2.100}=\frac{101}{200}\)
MA
2
BT
12 tháng 5 2017
\(C=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
Ta có: \(\frac{3}{4}=1-\frac{1}{4};\frac{8}{9}=1-\frac{1}{9};\frac{15}{16}=1-\frac{1}{16};...;\frac{9999}{10000}=1-\frac{1}{10000}\)
=> \(C=1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+1-\frac{1}{10000}\)
=> \(C=\left(1+1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)(99 chữ số 1)
=> \(C=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
Ta lại có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\); \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\); \(\frac{1}{4^2}>\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\); ...;\(\frac{1}{100^2}< \frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1-\frac{1}{100}=\frac{99}{100}< 1\)
=> C > 99-1
=> C > 98
12 tháng 5 2017
Ta có :C=
\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{100^2-1}{100^2}\)
\(=\frac{2^2}{2^2}+\frac{3^2}{3^2}+...+\frac{100^2}{100^2}-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
\(=99-\)\(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Mà \(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)l<\(\frac{100}{101}\)(tự tính)
Suy ra C> 98(đpcm)
NY
1
WH
4 tháng 5 2018
Trả lời
Ta có:
\(C=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(\Rightarrow C=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(\Rightarrow C=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)(99 chữ số 1)
\(\Rightarrow C=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
Ta lại có:
\(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Đặt D\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow D< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow D< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow D< 1-\frac{1}{100}\)
\(\Rightarrow D< \frac{99}{100}< 1\)
\(\Rightarrow C>99-1\)
\(\Rightarrow C>98\)
Vậy C>98 (đpcm)
NV
0
TT
1
HD
27 tháng 1 2020
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.
5 tháng 3 2018
Đặt \(A=\frac{3}{4}+\frac{8}{9}+..........+\frac{9999}{10000}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+..........+\left(1-\frac{1}{10000}\right)\)
\(=1-\frac{1}{2^2}+1-\frac{1}{3^2}+...........+1-\frac{1}{100^2}\)
\(=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{100^2}\right)\)\(>99-\left(\frac{1}{1.2}+\frac{1}{2.3}+..........+\frac{1}{99.100}\right)\)
\(=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99-\left(1-\frac{1}{100}\right)=99-1+\frac{1}{100}=98+\frac{1}{100}>98\)
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