tìm D biết D=\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right).....\left(\frac{1}{100^2}-1\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
22 tháng 3 2018
\(D=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(D=\left(\frac{3}{2\cdot2}\right)\left(\frac{8}{3\cdot3}\right)\left(\frac{15}{4\cdot4}\right)...\left(\frac{9999}{100\cdot100}\right)\)
\(D=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(99\cdot101\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(100\cdot100\right)}\)
\(D=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)
\(D=\frac{1\cdot101}{100\cdot2}\)
\(=\frac{101}{200}\)
22 tháng 3 2018
\(D=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot\cdot\cdot\left(\frac{1}{100^2}-1\right)\)(có 50 thừa số nên tích đó là số dương)
\(\Rightarrow D=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\cdot\cdot\cdot\left(\frac{100^2-1}{100^2}\right)\)
\(D=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\cdot\cdot\frac{99\cdot101}{100^2}\)
\(D=\frac{101}{2\cdot100}=\frac{101}{200}\)
S
14 tháng 4 2019
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
S
14 tháng 4 2019
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)
31 tháng 7 2019
Đề câu C sai nhé, sửa: ... < 1/2
\(C=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\\ 3C=1+\frac{1}{3}+...+\frac{1}{3^{98}}\\ 3C-C=1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{99}}\\ 2C=1-\frac{1}{3^{99}}\\ C=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\left(đpcm\right)\)
Đề câu D sai nhé, sửa: ... > -1/2
\(D=\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)\cdot...\cdot\left(\frac{1}{100^2}-1\right)< \left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)\cdot...\cdot\left(\frac{1}{100}-1\right)\)
Mặt khác \(\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)\cdot...\cdot\left(\frac{1}{100}-1\right)\\ =\frac{-1}{2}\cdot\frac{-2}{3}\cdot\frac{-3}{4}\cdot...\cdot\frac{-99}{100}\\ =-\left(\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\right)\\ =\frac{-1}{100}\)
Mà \(\frac{1}{100}< \frac{1}{2}\Rightarrow\frac{-1}{100}>\frac{-1}{2}\)
Vậy \(D< \frac{-1}{2}\left(đpcm\right)\)
5 tháng 7 2016
\(D=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{100^2}\right).\)
\(D=-\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\frac{4^2-1}{4^2}\cdot...\cdot\frac{100^2-1}{100^2}.\)
\(D=-\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\frac{3\cdot5}{4^2}\cdot\frac{4\cdot6}{5^2}\cdot...\cdot\frac{98\cdot100}{99^2}\cdot\frac{99\cdot101}{100^2}=-\frac{1}{2}\cdot\frac{101}{100}=-\frac{101}{200}\)
14 tháng 8 2017
câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
S
28 tháng 4 2016
D= [(1-1/2)(1-1/3)...(1-1/25)]:[(1+1/2)(1+1/3)...(1+1/25)]
D= [1/2. 2/3. ... . 24/25]: [3/2. 4/3. ... . 26/25]
D= 1/25 : 2/26
D= 1/25 . 26/2= 13/25
Vậy D= 13/25
28 tháng 4 2016
\(D=\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{25}\right)\right]\)\(:\left[\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{25}\right)\right]\)
\(D=\left[\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{24}{25}\right]:\left[\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{26}{25}\right]\)
\(D=\frac{1.2.3...24}{2.3.4...25}:\frac{3.4.5...26}{2.3.4...25}\)
\(D=\frac{1}{25}:13\)
\(D=\frac{1}{325}\)
\(D=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}...........\frac{1-100^2}{100^2}\)
\(=\frac{\left(-1\right).3}{2^2}.\frac{\left(-2\right).4}{3^2}..............\frac{\left(-99\right).101}{100^2}\)
\(=\frac{\left(-1\right).\left(-2\right)........\left(-99\right)}{2.3..............100}.\frac{3.4..............101}{2.3..............100}\)
\(=\frac{-1}{100}.\frac{101}{2}=\frac{-101}{200}\)