Tính: \(E=\frac{3}{1.3.5}+\frac{3}{3.5.7}+..+\frac{3}{13.15.17}\)
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chứng tỏ rằng : A=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+....+\frac{36}{25.27.29}< 3\)
1
VM
6 tháng 2 2020
Ta có:
\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
\(\Rightarrow A=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)
\(\Rightarrow A=9.\left(\frac{1}{3}-\frac{1}{783}\right)\)
\(\Rightarrow A=9.\frac{1}{3}-9.\frac{1}{783}\)
\(\Rightarrow A=3-\frac{1}{87}\)
Vì \(3-\frac{1}{87}< 3.\)
\(\Rightarrow A< 3\left(đpcm\right).\)
Chúc bạn học tốt!
MT
11 tháng 7 2015
a)\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
=\(\frac{9.4}{1.3.5}+\frac{9.4}{3.5.7}+\frac{9.4}{5.7.9}+...+\frac{9.4}{25.27.29}\)
=\(9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
=\(9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
=\(9.\left(\frac{1}{3}-\frac{1}{27.29}\right)=9.\left(\frac{1}{3}-\frac{1}{783}\right)=9.\left(\frac{261}{783}-\frac{1}{783}\right)=9.\frac{260}{783}\)
=\(\frac{260}{87}\)
b)
ta có: \(3=\frac{261}{87}>\frac{260}{87}\)
vậy A<3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 3
\(\dfrac{3}{2.6}\) + \(\dfrac{3}{6.10}\) + \(\dfrac{3}{10.14}\)
= \(\dfrac{3}{4}\).(\(\dfrac{4}{2.6}\) + \(\dfrac{4}{6.10}\) + \(\dfrac{4}{10.14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}-\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\). \(\dfrac{3}{7}\)
= \(\dfrac{9}{28}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 3
B = \(\dfrac{4}{1.3.5}\) + \(\dfrac{4}{3.5.7}\) + \(\dfrac{4}{5.7.9}\)
B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{3.5}\) + \(\dfrac{1}{3.5}\) - \(\dfrac{1}{5.7}\) + \(\dfrac{1}{5.7}\) - \(\dfrac{1}{7.9}\)
B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{7.9}\)
B = \(\dfrac{1}{3}\) - \(\dfrac{1}{63}\)
B = \(\dfrac{20}{63}\)
ML
26 tháng 6 2015
Áp dụng: \(\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}\)
\(\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}
8 tháng 8 2016
Xét tử số có dạng : \(\frac{1}{\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)}=\frac{1}{4}\left[\frac{1}{\left(2n+1\right)\left(2n+2\right)}-\frac{1}{\left(2n+2\right)\left(2n+3\right)}\right]\) với \(n\in N\)
Ta có : \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2005.2007.2009}\)
\(=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\frac{1}{4}.\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\frac{1}{4}\left(\frac{1}{5.7}-\frac{1}{7.9}\right)+...+\frac{1}{4}\left(\frac{1}{2005.2007}-\frac{1}{2007.2009}\right)\)
\(=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2005.2007}-\frac{1}{2007.2009}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{2007.2009}\right)\)
Xét mẫu số có dạng : \(\frac{1}{\left(2n+1\right)\sqrt{2n+3}+\left(2n+3\right)\sqrt{2n+1}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)
\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{\sqrt{2n+1}.\sqrt{2n+3}\left[\left(2n+3\right)-\left(2n+1\right)\right]}=\frac{1}{2}.\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với \(n\in N\)
Áp dụng : \(\frac{1}{1\sqrt{3}+3\sqrt{1}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{2007\sqrt{2009}+2009\sqrt{2007}}\)
\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{2007}}-\frac{1}{\sqrt{2009}}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{2009}}\right)\)
Suy ra : \(M=\frac{\frac{1}{4}\left(\frac{1}{3}-\frac{1}{2007.2009}\right)}{\frac{1}{2}\left(1-\frac{1}{\sqrt{2009}}\right)}\)
Tới đây bài toán đã gọn hơn , bạn tự tính nhé :)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
25 tháng 4 2017
\(\frac{C}{9}=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{21.23.25}+\frac{4}{23.25.27}.\)
\(\frac{C}{9}=\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+\frac{9-5}{5.7.9}+...+\frac{25-21}{21.23.25}+\frac{27-23}{23.25.27}\)
\(\frac{C}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{21.23}-\frac{1}{23.25}+\frac{1}{23.25}-\frac{1}{25.27}\)
\(\frac{C}{9}=\frac{1}{3}-\frac{1}{25.27}\Rightarrow C=\frac{9\left(25.9-1\right)}{25.27}=\frac{25.9-1}{25.3}=3-\frac{1}{25.3}< 3\)
PT
7 tháng 11 2017
\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{1997.1999}-\frac{1}{1999.2001}\)
\(=\frac{1}{1.3}-\frac{1}{1999.2001}\)
Bạn tính kết quả nhé
NM
3
BM
19 tháng 3 2016
Ta có :
2K = \(\frac{4}{1.3.5}+\frac{4}{3.5.7}+...............+\frac{4}{99.101.103}=\)\(\frac{1}{1.3}-\frac{1}{3.5}+\)\(+\frac{1}{3.5}-\frac{1}{5.7}+..................+\frac{1}{99.101}-\frac{1}{101.103}=\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+................+\frac{1}{101}-\frac{1}{103}=1-\frac{1}{103}=\frac{102}{103}\)
=> K= \(\frac{102}{103}:2=\frac{51}{103}\)
\(2E=\frac{6}{1.3.5}+\frac{6}{3.5.7}+...+\frac{3}{13.15.17}\)
\(2E=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{13.15}-\frac{1}{15.17}\)
\(2E=\frac{1}{1.3}-\frac{1}{15.17}\)
\(2E=\frac{1}{15}-\frac{1}{255}\)
\(\Rightarrow2E=\frac{16}{255}\)
\(\Rightarrow E=\frac{8}{255}\)
Dòng thứ 4 là sao