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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.
Ta có :
\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(A=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
\(A=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>99\)\(\left(1\right)\)
gọi B là biểu thức trong ngoặc
Lại có :
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B< 1-\frac{1}{100}< 1\)
\(\Rightarrow A=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>99-\left(1-\frac{1}{100}\right)>98\)
\(\Rightarrow A>98\)\(\left(2\right)\)
từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\)\(98< A< 99\)
vậy A không phải là số tự nhiên
phần bạn đánh dấu (1) thì A<99 vì A= 99 trừ đi một số mà
Ta có:
\(\dfrac{n^2-1}{n^2}=1-\dfrac{1}{n^2}>1-\dfrac{1}{\left(n-1\right)n}\)
Áp dụng:
\(C=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+\dfrac{4^2-1}{4^2}+...+\dfrac{100^2-1}{100^2}\)
\(C>1-\dfrac{1}{1.2}+1-\dfrac{1}{2.3}+1-\dfrac{1}{3.4}+...+1-\dfrac{1}{99.100}\)
\(C>99-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(C>99-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(C>99-\left(1-\dfrac{1}{100}\right)\)
\(C>98+\dfrac{1}{100}>98\) (đpcm)
\(C=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{9999}{10000}\)
\(C=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{99\cdot101}{100\cdot100}\)
\(C=\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\frac{3\cdot4\cdot5\cdot...\cdot101}{2\cdot3\cdot4\cdot...\cdot100}\)
\(C=\frac{1}{100}\cdot\frac{101}{2}\)
\(C=\frac{101}{200}\)
\(C=\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x......x\frac{9999}{10000}\)
\(C=\frac{1.3}{2^2}x\frac{2.4}{3^2}x\frac{3.5}{4^2}x....x\frac{99.101}{100^2}\)
\(C=\frac{1.3.2.4.3.5.......99.101}{2^2.3^2.4^2.......100^2}\)
\(C=\frac{1.2.3.......99}{2.3.4....100}x\frac{3.4.5.....101}{2.3.4......100}\)
\(C=\frac{1}{100}.\frac{101}{2}=\frac{1.101}{100.2}=\frac{101}{200}\)
Ủng hộ mk nha!!!!
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 \(B=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)>99-1=98\)
= 3/22 + 8/32 + 15/42 + ... + 9999/1002
= 1.3/2.2 + 2.4/3.3 + 3.5/4.4 + .... + 99.101/100.100
\(=\frac{1.3.2.4.3.5.4.6...99.101}{2^2.3^2....100^2}=\frac{1.2.3^2.4^2...99^2.100.101}{2^2.3^2....100^2}=\frac{1.2.101}{2^2.100}=\frac{101}{200}\)
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)
Ta có : \(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}=\frac{4-1}{4}+\frac{9-1}{9}+\frac{16-1}{16}+...+\frac{10000-1}{10000}\)
\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{100^2-1}{100^2}\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\left(99\text{ số hạng 1}\right)\)
\(=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>99-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)
\(=99-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)=99-\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=99-\frac{99}{202}>99-\frac{1}{2}=98,5\)
=> A > 98,5
=> A > 98
\(C=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
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
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)
Ta có: \(\frac{3}{4}=1-\frac{1}{4}=1-\frac{1}{2^2}\); \(\frac{8}{9}=1-\frac{1}{9}=1-\frac{1}{3^2}\)
\(\frac{15}{16}=1-\frac{1}{16}=1-\frac{1}{4^2}\); ...; \(\frac{9999}{10000}=1-\frac{1}{10000}=1-\frac{1}{100^2}\)
=> \(C=\left(1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
=> \(C=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)=99-B\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=> \(B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)
=> A > 99-1 = 98
=> B > 98
B ở đâu ra thế