Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)
\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)
\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)
\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)
Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B
\(\left(1+\dfrac{1}{2}\right)+\left(1+\dfrac{1}{2^2}\right)+...+\left(1+\dfrac{1}{2^{50}}\right)\)
= \(\left(1+1+1+...+1\right)+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)(50 số 1 )
= \(50+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)
A =\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\)
⇒ 2A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\)
⇒ 2A - A =\(1-\dfrac{1}{2^{50}}\)
=50+1-\(\dfrac{1}{2^{50}}\)=51-\(\dfrac{1}{2^{50}}>3\)
`A = 3/4 xx 8/9 xx ... xx 99/100`
`= (1xx3)/(2xx2) xx (2xx4)/(3xx3) xx ... xx (9xx11)/(10xx10)`
`= (1xx2xx3xx ... xx 9)/(2xx3xx...xx10) xx (3xx4xx5xx...xx 11)/(2xx3xx4xx...xx 10)`
`= 1/10 xx 11`
`= 11/10`.
Ta có: `11/10 > 1`
`11/19 < 1`.
`=> A > 11/19`.
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
\(\left(\dfrac{1}{2^2}-1\right)\times\left(\dfrac{1}{3^2-1}\right)\times\left(\dfrac{1}{4^2}-1\right)\times...\times\left(\dfrac{1}{100^2}-1\right)\)
\(=\dfrac{3}{2^2}\times\dfrac{8}{3^2}\times\dfrac{15}{4^2}\times...\times\dfrac{100^2-1}{100^2}\)
\(=\dfrac{1\times3}{2\times2}\times\dfrac{2\times4}{3\times3}\times\dfrac{3\times5}{4\times4}\times...\times\dfrac{99\times101}{100\times100}\)
\(=\dfrac{1\times2\times3\times...\times99}{2\times3\times4\times...\times100}\times\dfrac{3\times4\times5\times...\times101}{2\times3\times4\times...\times100}\)
\(=\dfrac{1}{100}\times\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
\(\left(\dfrac{1}{2^2}-1\right)\cdot\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\dfrac{-3}{4}\cdot\dfrac{-8}{3}\cdot...\cdot\dfrac{-9999}{10000}\)
\(=\dfrac{1\cdot\left(-3\right)}{2\cdot2}\cdot\dfrac{2\cdot\left(-4\right)}{3\cdot3}\cdot...\cdot\dfrac{99\cdot\left(-101\right)}{100\cdot100}\)
\(=\dfrac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\dfrac{\left(-3\right)\cdot\left(-4\right)\cdot...\cdot\left(-101\right)}{2\cdot3\cdot4\cdot...\cdot100}\)
Ở tử số phân số bên phải có số thừa số là: \(101-3+1=99\)
99 là số lẻ nên tử số vế phải sẽ cho ra số âm.
\(=\dfrac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot\left(-101\right)}{2\cdot3\cdot4\cdot...\cdot100}\)
\(=\dfrac{1\cdot\left(-101\right)}{100\cdot2}\)
\(=\dfrac{-101}{200}\)
Lời giải:
Xét thừa số tổng quát:
\(1-\frac{1}{1+2+...+n}=1-\frac{1}{\frac{n(n+1)}{2}}=1-\frac{2}{n(n+1)}=\frac{n(n+1)-2}{n(n+1)}\)
\(=\frac{n^2-1+n-1}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}\)
Do đó:
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{99.102}{100.101}\)
\(=\frac{(1.2.3...99)(4.5.6...102)}{(2.3.4...100)(3.4.5..101)}=\frac{1}{100}.\frac{102}{3}=\frac{102}{300}\)
Ta có: \(1-\dfrac{1}{k^2}=\dfrac{k^2-1}{k^2}=\dfrac{\left(k+1\right)\left(k+1\right)}{k^2}\) nên:
\(A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{99.101}{100^2}=\dfrac{1.3.2.4.....99.101}{2^2.3^2.4^2....100^2}=\dfrac{1.2.3.....99}{2.3.4.5....99.100}.\dfrac{3.4.5.....101}{2.3.4.5.....100}=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)
Chúc bạn học tốt !!!
\(D=\dfrac{\left(2!\right)^2}{1^2}+\dfrac{\left(2!\right)^2}{3^2}+\dfrac{\left(2!\right)^2}{5^2}+...+\dfrac{\left(2!\right)^2}{2015^2}\)
\(D=\left(2!\right)^2\left(\dfrac{1}{3^2}+\dfrac{1}{5^2}+...+\dfrac{1}{2015^2}\right)\)
Xét số hạng tổng quát dạng: \(\dfrac{1}{\left(2n+1\right)^2}\) với \(n\in N\ge1\)
Ta có: \(\left(2n+1\right)^2-2n\left(2n+1\right)=1>0\)
\(\Rightarrow\left(2n+1\right)^2>2n\left(2n+1\right)\Rightarrow\dfrac{1}{\left(2n+1\right)^2}< \dfrac{1}{2n\left(2n+1\right)}\)
Do đó: \(\left\{{}\begin{matrix}\dfrac{1}{3^2}< \dfrac{1}{2.4}\\\dfrac{1}{5^2}< \dfrac{1}{4.6}\\....\\\dfrac{1}{2015^2}< \dfrac{1}{2014.2016}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}...+\dfrac{1}{2015^2}< 1+\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{2014.2016}\)
\(\Leftrightarrow\dfrac{D}{\left(2!\right)^2}< 1+\dfrac{1}{2.4}+\dfrac{1}{4.6}+..+\dfrac{1}{2014.2016}\)
\(\Leftrightarrow D< 4\left(1+\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{2014.2016}\right)\)
\(\Leftrightarrow D< 4+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{1007.1008}\)
\(\Leftrightarrow D< 4+\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+...+\dfrac{1008-1007}{1007.1008}\)
\(\Leftrightarrow D< 4+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{...1}{1007}-\dfrac{1}{1008}\)
\(\Leftrightarrow D< 5-\dfrac{1}{1008}< 5< 6\)