Cho tổng A gồm 2014 số hạng: A = \(\frac{1}{19}\)+ \(\frac{2}{19^2}\)+\(\frac{3}{19^3}\)+...+\(\frac{2014}{19^{2014}}\)
Hãy so sánh A2013 và A2014.
Bác nào giả dc e hậu tạ 3tik
~ARMY~
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Câu 5. (1,0 điểm)
Cho tổng A gồm 2014 số hạng: A = \(\frac{1}{19}+\frac{2}{19^2}+\frac{3}{19^3}+..........+\frac{2014}{19^{2014}}\)
Hãy so sánh A2013 và A2014.
Cho tổng A gồm 2014 số hạng.
\(A=\frac{1}{19}+\frac{2}{19^2}+\frac{3}{19^3}+...+\frac{2014}{19^{2014}}\)
Tính A
\(A=\frac{1}{19}+\frac{2}{19^2}+...+\frac{2014}{19^{2014}}\)
\(\Rightarrow19A=1+\frac{2}{19}+\frac{3}{19^2}+...+\frac{2014}{19^{2013}}\)
\(\Rightarrow19A-A=\left(1+\frac{2}{19}+\frac{3}{19^2}+...+\frac{2014}{19^{2013}}\right)-\left(\frac{1}{19}+\frac{2}{19^2}+...+\frac{2014}{19^{2014}}\right)\)
\(\Rightarrow18A=1+\left(\frac{1}{19}+\frac{1}{19^2}+...+\frac{1}{19^{2013}}\right)-\frac{2014}{19^{2014}}\)
\(\Rightarrow18A=1+\frac{1-\frac{1}{19^{2013}}}{18}-\frac{2014}{19^{2014}}\)
\(\Rightarrow A=\frac{1+\frac{1-\frac{1}{19^{2013}}}{18}-\frac{2014}{19^{2014}}}{18}\)
Vậy...
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Cho tổng A gồm 2016 số hạng A=\(\frac{1}{19^1}+\frac{2}{19^2}_{ }+\frac{3}{19^3}+..................+\frac{n}{19^n}+.....+\frac{2016}{19^{2016}}\)
Hãy so sánh A^2016 và A^2015
Ai giải được cho 100 tick
Không cần giải cũng biết đáp án:
Nếu A là số dương thì A^2016>A^2015
Nếu A là số âm thì A^2016 là số dương , A^2015 là số âm nên chắc chắn A^2016>A^2015
k nha
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So sánh \(A=\frac{19^{2015}+3}{19^{2016}+3}vàB=\frac{19^{2014}+3}{19^{2015+3}}\)
Câu 1 :
So sánh:
A = \(\frac{-9}{10^{2013}}\)+\(\frac{-19}{10^{2014}}\)và B=\(\frac{-9}{10^{2014}}\)+\(\frac{-19}{10^{2013}}\)
quy dong ca A va B ta dc :
\(A=\frac{-109}{10^{2014}}\)
\(B=\frac{-199}{10^{2014}}\)
\(\Rightarrow A>B\)
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dễ thôi
ta có :A=-9/10^2013+-19/10^2014=-9/10^2013+-9/10^2014+-10/10^2014
B=-9/10^2014+-19/10^2013=-9/10^2014+-9/10^2013+-10/10^2013
nhìn nhé :cả A và B đều có các số hạng :-9/10^2013 và-9/10^2014
mà -10/10^2014<-10/10^2013
=>A<B
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1/Tính
a) \(\left(\frac{7}{2}+\frac{17}{3}\right).\frac{8}{19}-\left(\frac{5}{2}+\frac{14}{3}\right).\frac{8}{19}\)
b)\(\frac{3^{2014}.8^{19}}{6^{60}.3^{1955}}\)
a) Cho tổng gồm 2014 số hạng
S= \(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2014}{4^{2014}}\)
CMR S<1
\(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+....+\frac{2014}{4^{2014}}\)
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2014}{4^{2013}}\)
\(4S-S=\left(1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2014}{4^{2013}}\right)-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2014}{4^{2014}}\right)\)
\(3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
\(12S=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}-\frac{2014}{4^{2013}}\)
\(12S-3S=\left(4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}-\frac{2014}{4^{2013}}\right)-\left(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\right)\)
\(9S=4-\frac{2014}{4^{2013}}-\frac{1}{4^{2013}}+\frac{2014}{4^{2014}}\)
\(9S=4-\frac{4028}{4^{2014}}-\frac{4}{4^{2014}}+\frac{2014}{4^{2014}}\)
\(9S=4-\frac{2010}{4^{2014}}< 4\)
\(\Rightarrow9S< 4\)
\(\Rightarrow S< \frac{4}{9}< 1\)(đpcm)
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Ta có :
\(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\)( 1 )
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2014}{4^{2013}}\)( 2 )
Lấy ( 2 ) - ( 1 ) ta được :
\(3S=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
gọi \(B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)( 3 )
\(4B=4+1+\frac{1}{4}+...+\frac{1}{4^{2012}}\) ( 4 )
Lấy ( 4 ) - ( 3 ) ta được :
\(3B=4-\frac{1}{4^{2013}}\)
\(\Rightarrow B=\frac{4-\frac{1}{4^{2013}}}{3}=\frac{4}{3}-\frac{1}{4^{2013}.3}\)
\(\Rightarrow3S=\frac{4}{3}-\frac{1}{4^{2013}.3}-\frac{2014}{4^{2014}}\)
\(\Rightarrow S=\frac{\frac{4}{3}-\frac{1}{4^{2013}.3}-\frac{2014}{4^{2014}}}{3}=\frac{4}{9}-\frac{1}{4^{2013}.9}-\frac{2014}{4^{2014}.3}< \frac{4}{9}< 1\)
vậy \(S< 1\)
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So sánh : a) 17 \(^{19}\)+ 17\(^{17}\)và 2.17\(^{18}\)
b) \(\frac{3^{2015}-1}{3^{2014}-1}\)và \(\frac{3^{2014}-1}{3^{2015}-1}\)
Cho tổng gồm 2014 số hạng, \(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2014}{4^{2014}}\). Chứng minh rằng \(S< \frac{1}{2}\).
Xét \(4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+\dfrac{4}{4^3}+...+\dfrac{2014}{4^{2013}}\)
=> \(3S=4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2014}{4^{2013}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+...+\dfrac{2014}{4^{2014}}\right)\)
=> \(3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}-\dfrac{2014}{4^{2014}}< 1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\)
Đặt \(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\)
=> \(4A=4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\)
=> \(3A=4A-A=\left(4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\right)\)
=> \(3A=4-\dfrac{1}{4^{2013}}< 4\)
=> \(A< \dfrac{4}{3}\)
=> \(3S< \dfrac{4}{3}\)
=> \(S< \dfrac{4}{9}< \dfrac{1}{2}\)
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\(4S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}\)
\(4S-S=3S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2014}{4^{2014}}\right)\)
\(3S=1+\left(\frac{2}{4}-\frac{1}{4}\right)+\left(\frac{3}{4^2}-\frac{2}{4^2}\right)+......+\left(\frac{2014}{4^{2013}}-\frac{2013}{4^{2013}}\right)-\frac{2014}{4^{2014}}\)
\(3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+.....+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
đặt \(A=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2023}}\)
\(4A-A=4+1+\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{2022}}-\left(1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2023}}\right)\)
\(3A=4-\frac{1}{4^{2023}}\)
\(A=\frac{4}{3}-\frac{1}{3.4^{2023}}\)
\(\Rightarrow3S=\frac{4}{3}-\frac{1}{3.4^{2023}}-\frac{2014}{4^{2024}}\)
\(\Rightarrow S=\frac{4}{9}-\frac{1}{9.4^{2023}}-\frac{2014}{3.4^{2024}}\)
do \(\frac{4}{9}< \frac{4}{8}=\frac{1}{2}\)
\(\Rightarrow S=\frac{4}{9}-\frac{1}{9.4^{2023}}-\frac{2014}{3.4^{2024}}< \frac{4}{8}=\frac{1}{2}\left(đpcm\right)\)