Cho \(S=\frac{3}{4}+\frac{15}{16}+\frac{24}{25}+\frac{35}{36}+...+\frac{9999}{10000}\). Chứng minh rằng : \(S\notinℕ\)
Cho \(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}+...+\frac{9999}{10000}\). Chứng minh rằng : S\(\notin\)N.
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.
Cho S = \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\). Chứng minh rằng s không phải là số nguyên
@Nguyễn Văn Đạt nhờ a nha
\(Ta\) \(có\) :
\(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{100^2-1}{100^2}\)
\(=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
\(Đặt\) \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Do A > 0 nên S < 99 (1)
Do A\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}\)
Suy ra \(S=99-A>99-\left(1-\frac{1}{100}\right)\)
\(\Rightarrow S>98+\frac{1}{100}\Rightarrow S>98\) (2)
Lập luận ra điều phải chứng minh
\(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(\Rightarrow S=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(\Rightarrow S=1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+1-\frac{1}{10000}\)
\(\Rightarrow S=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
\(\Rightarrow S=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)< 99.\)
\(\Rightarrow S< 99\) (1).
Đặt \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có:
\(\left\{{}\begin{matrix}\frac{1}{2^2}< \frac{1}{1.2}\\\frac{1}{3^2}< \frac{1}{2.3}\\....\\\frac{1}{100^2}< \frac{1}{99.100}\end{matrix}\right.\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}\)
Vì \(1-\frac{1}{100}< 1.\)
\(\Rightarrow A< 1.\)
\(\Rightarrow S>99-1\)
\(\Rightarrow S>98\) (2).
Từ (1) và (2) \(\Rightarrow98< S< 99.\)
\(\Rightarrow S\) không phải là số nguyên (đpcm).
Chúc bạn học tốt!
B=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{9999}{10000}=?\)
\(B=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right).\left(4.6\right)...\left(99.101\right)}{2^2.3^2.4^2.5^2...100^2}=\frac{\left(1.2.3.4...99\right).\left(3.4.5.6...101\right)}{\left(2.3.4.5...100\right)\left(2.3.4.5...100\right)}=\frac{1.101}{100.2}=\frac{101}{200}\)
B = \(\frac{1.3}{2^2}.\frac{2.4}{3^2}\frac{3.5}{4^2}\frac{4.6}{5^2}...\frac{99.101}{100^2}=\frac{1.3.2.4.3.5.4.6...99.101}{2.2.3.3.4.4.5.5...100.100}\)
=\(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
Vật B = \(\frac{101}{200}\)
đúng cái đi
Chứng tỏ rằng \(\frac{3}{4}\)+\(\frac{8}{9}\)+\(\frac{15}{16}\)+\(\frac{24}{25}\)+....+\(\frac{9999}{10000}\)không phải là số nguyên
Đặt A =\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
\(=99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)\)
Đặt B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
>\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{2}-\frac{1}{101}=\frac{99}{202}\)
Khi đó A = \(99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)< 99-\frac{99}{202}\approx98,5\)
=> A < 98,5 (1)
Lại có B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
Khi đó A =\(99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)>99-\frac{99}{100}=98,01\)
=> A > 98,01 (2)
Từ (1)(2) => 98,01 < A < 98,5
=> A không là số nguyên
Cho C=\(\frac{3}{4}\)+\(\frac{8}{9}\)+\(\frac{15}{16}\)+.....+\(\frac{9999}{10000}\)
Chứng minh rằng C>98
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
Tính giá trị của A biết:
1/ \(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{9999}{10000}\)
2/ \(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{3599}{3600}\)
mọi nguời cứ tính đi đuợc câu nào tôi cũng tick cho
Chứng minh \(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)không phải là số tự nhiên
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
Cho S=\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}\)\(+...+\frac{2499}{2500}\). Chứng tỏ rằng S không phải là số tự nhiên.
Help me!!!
Cho \(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}+...+\frac{2499}{2500}\)
Chứng minh S ko phải là số tự nhiên.
GIẢI CHI TIẾT NHÉ.MÌNH CẢM ƠN!!!!!
Ta có :
\(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{5000}\)
\(S=1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+1-\frac{1}{5000}\)
\(S=\left(1+1+1+...+1\right)-\left(\frac{1}{4}++\frac{1}{9}+\frac{1}{16}+...+\frac{1}{5000}\right)\)
\(S=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)< 49\)\(\left(1\right)\)
Lại có :
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
\(\Rightarrow\)\(-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>-1\)
\(\Rightarrow\)\(S=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>49-1=48\)\(\left(2\right)\)
Từ (1) và (2) suy ra :
\(48< S< 49\)
Vậy S không là số tự nhiên
Chúc bạn học tốt ~
\(S=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+...+\left(1-\frac{1}{2500}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
\(=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< 49\left(1\right)\)
Có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
\(\Rightarrow-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)>-1\)
\(\Rightarrow A=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)>49-1=48\)(2)
Từ (1) và (2) => 48<A<49
Vậy S không phải là stn
Ta có: \(\frac{3}{4}=1-\frac{1}{4}=1-\frac{1}{2^2}\); \(\frac{8}{9}=1-\frac{1}{9}=1-\frac{1}{3^2}\);...; \(\frac{2499}{2500}=1-\frac{1}{2500}=1-\frac{1}{50^2}\)
=> \(S=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}\right)\)
Xét \(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
=> \(M< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}=\frac{49}{50}\)
Lại có: \(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\)
=> \(M>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}=\frac{1}{2}-\frac{1}{51}=\frac{49}{102}\)
=> \(\frac{49}{102}< M< \frac{49}{50}< 1\)
Như vậy, M không phải là số tự nhiên => S=49-M không phải là số tự nhiên