Chứng tỏ : 1/1+3 + 1/1+3+5 + 1/1+3+5+7+...+1/1+3+5+7+...+2017 < 3/4
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
27 tháng 5 2017
Có \(A=\dfrac{1}{1+3}+\dfrac{1}{1+3+5}+...+\dfrac{1}{1+3+5+...+2017}\)
\(\Rightarrow A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{1+3+...+2017}\)
\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2017^2}\)
Ta thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{4}\)
\(\dfrac{1}{3^2}< \dfrac{1}{3.2}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
.................
\(\dfrac{1}{2017^2}< \dfrac{1}{2016.2017}\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2016.2017}\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2016}-\dfrac{1}{2017}\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{2017}\)
\(\Rightarrow A< \dfrac{3}{4}-\dfrac{1}{2017}\)
\(\Rightarrow A< \dfrac{3}{4}\)
Vậy \(A< \dfrac{3}{4}\).
27 tháng 5 2017
Có \(\dfrac{1}{1+3}\) + \(\dfrac{1}{1+3+5}\) +...+ \(\dfrac{1}{1+3+...+2017}\)
= \(\dfrac{1}{2^2 }\)+\(\dfrac{1}{3^2}\) + ... +\(\dfrac{1}{2017^2}\)
Lại có :
\(\dfrac{1}{2^2}\) = \(\dfrac{1}{4} \)
\(\dfrac{1}{3^2}\) <\(\dfrac{1}{2.3}\)
...
\(\dfrac{1}{2017^2}\) <\(\dfrac{1}{2016.2017}\)
\(\Rightarrow \) A< \(\dfrac{1}{4} \) +\(\dfrac{1}{2.3}\)+... +\(\dfrac{1}{2016.2017}\)
A<\(\dfrac{1}{4} \)+\(\dfrac{1}{2}\)- \(\dfrac{1}{3}\) +...+\(\dfrac{1}{2016}- \dfrac{1}{2017}\)
A< \(\dfrac{1}{4} \)+\(\dfrac{1}{2}\) -\(\dfrac{1}{2017}\)
A<\(\dfrac{3}{4}\) -\(\dfrac{1}{2017}\)
\(\Rightarrow\)A<\(\dfrac{3}{4}\) (đpcm)
chúc bạn học tốt !!!
21 tháng 4 2017
A=1/(1+3)+1/(1+3+5)+1/(1+3+5+7)+...+1/(1+3+5+7+...+2017)
A=1/2^2+1/3^2+1/4^2+...+1/1009^2
2A=2/2^2+2/3^2+2/4^2+...+2/1009^2
Ta co :(x-1)(x+1)=(x-1)x+x-1=x^2-x+x-1=x^2-1<x^2
suy ra 2A<2/(1*3)+2/(3*5)+2/(5*7)+...+2/(1008*1010)
suy ra 2A <1-1/3+1/3-1/5+1/5-1/7+...+1/1008-1/1010
suy ra 2A<1-1/1010
suy ra 2A<2009/2010<1<3/2
suy ra 2A <3/2
suy ra A <3/4 (dpcm)
nho k cho minh voi nha
HH
6 tháng 5 2017
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{1009^2}\)
Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{4};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{1009^2}< \dfrac{1}{1008.1009}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{1009^2}< \dfrac{1}{4}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{1008.1009}\)\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1008}-\dfrac{1}{1009}\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{1009}\)
\(\Rightarrow A< \dfrac{3}{4}-\dfrac{1}{1009}\)
\(\Rightarrow A< \dfrac{3}{4}\left(đpcm\right)\)
LT
1
9 tháng 4 2019
A=1/2^2+1/3^2+....+1/1009^2
2A=2/2^2+2/3^2+...+2/1009^2
Ta có : (x-1).(x+1)=(x-1).x+x-1=x^2-x+x-1=x^2-1<x^2
2A<2/1.3+2/3.5+2/5.7+...+2/1008.10010
2A<1-1/3+1/3-1/5+...+1/1008-1/1010
2A<1-1/1010
2A<1009/1010<1<3/2
2A<3/2
A<3/4
ĐPCM
Nhớ cho mình nha!
25 tháng 6 2018
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
\(\Rightarrow A=\frac{5}{11}\)
25 tháng 6 2018
\(2B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}\Rightarrow B=\frac{1009}{2019}\)
\(\frac{2}{7}C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}\Rightarrow C=\frac{2018}{2019}:\frac{2}{7}=\frac{7063}{2019}\)
\(VT=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{1018081}=\)
\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1008.1009}=\)
\(=\frac{1}{4}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{1009-1008}{1008.1009}=\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1008}-\frac{1}{1009}=\frac{3}{4}-\frac{1}{1009}< \frac{3}{4}\)