s=1/1.3+1/3.5+...+1[2n-1][2n+1]+...+25,5.257
ai biết bài này giải lẹ lên được không
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
DT
Tính S = 1.3/3.5 + 2.4/5.7 + 3.5/7.9 + ... + ( n-1)( n+1) / (2n-1)(2n+1) + ... + 1002.1004/2005.2007
2
KN
13 tháng 8 2020
\(S=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)
\(\Rightarrow S=\frac{\left(2-1\right)\left(2+1\right)}{\left(2.2-1\right)\left(2.2+1\right)}+\frac{\left(3-1\right)\left(3+1\right)}{\left(3.2-1\right)\left(3.2+1\right)}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}\)
\(+..+\frac{\left(1003-1\right)\left(1003+1\right)}{\left(1003.2-1\right)\left(1003.2+1\right)}\)
\(\Rightarrow S=\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}\right)+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{3.2-1}-\frac{1}{3.2+1}\right)+...\)
\(+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+...+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{1003.2-1}-\frac{1}{1003.2+1}\right)\)
\(\Rightarrow S=1002.\frac{1}{4}-1002.\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}+\frac{1}{3.2-1}-...-\frac{1}{1003.2+1}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{2007}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}.\frac{668}{2007}\)
\(\Rightarrow S=\frac{501}{2}-\frac{27889}{223}\)
\(\Rightarrow S=125,4372197\)
\(\)
NH
0
26 tháng 12 2018
a) Đặt B= 1/1.3 + 1/3.5 + 1/5.7 + .....+ 1/19.21
Ta có: 2B= 2/1.3 + 2/3.5 + 2/5.7 + ....+ 2/19.21
= 1- 1/3 + 1/3-1/5 + 1/5-1/7 +....+ 1/19-1/21
= 1-1/21 = 20/21
=> B= 20/21 : 2 => B= 10/21
b) Như trên, ta có: 2A= 1- (1/2n + 1) => A=( 1-1/2n+1).1/2
=> A= 1/2- 1/2n+1
=> A< 1/2 ( đpcm )
TP
1
DN
0
ND
27 tháng 2 2018
\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\) \(\Rightarrow2S=1-\dfrac{1}{2n+1}\)
\(\Rightarrow S=\dfrac{n}{2n+1}\)
27 tháng 2 2018
Ta có : \(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
ta được \(\dfrac{1}{1.3}=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right);\dfrac{1}{3.5}=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right);\dfrac{1}{5.7}=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\)
\(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\) vậy \(S=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)=\dfrac{n}{2n+1}\)
DD
Đoàn Đức Hà
Giáo viên
25 tháng 2 2021
\(A=1.3+3.5+5.7+...+\left(2n+1\right)\left(2n+3\right)\)
\(6A=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+\left(2n+1\right)\left(2n+3\right)\left(2n+5-2n+1\right)\)
\(6A=3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)\)
\(+\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)\)
\(6A=\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)+3\)
\(A=\frac{\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)+3}{6}\)
3 tháng 12 2017
2A = 2/1.3+2/3.5+....+2/(2n-1).(2n+1)
= 1-1/3+1/3-1/5+.....+1/2n-1 - 1/2n+1
= 1-1/2n+1 < 1
=> A < 1/2
=> ĐPCM
k mk nha
DN
2
10 tháng 1 2018
Đặt A = 1/1.3 + 1/3.5 + 1/5.7 +........+ 1/(2n - 1)(2n + 1)
2.A = 2/1.3 + 2/3.5 + 2/5.7 +........+ 2/(2n - 1)(2n + 1)
2.A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/(2n - 1) - 1/(2n + 1)
2.A = 1 - 1/(2n + 1) = 2n/(2n + 1)
Vậy A = n/(2n + 1)