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OD
3
OD
1
QD
2
29 tháng 11 2021
\(\Rightarrow3B=3^2+3^3+3^4+...+3^{101}\\ \Rightarrow3B-B=3^2+3^3+...+3^{101}-3-3^2-3^3-...-3^{100}\\ \Rightarrow2B=3^{101}-3\\ \Rightarrow B=\dfrac{3^{101}-3}{2}\)
NA
29 tháng 11 2021
B = 31 + 32 + 33 + .... + 399 + 3100
3B = 3(31 + 32 + 33 + ..... + 399 + 3100)
3B = 32 + 33 + 34 +...... + 3100 + 3101
3B - B = 2B = (32 + 33 + 34 + .... + 3100 + 3101) - ( 31 + 32 + 33 + .... + 3100)
2B = (32 - 32) + (33 - 33) +.....+ ( 3100 - 3100) + ( 3101 - 1)
2B = 0 + 0 + 0 + ..... +0 + 3101 - 1
2B = 3101 - 1
B = (3101 - 1) : 2
DT
2
6 tháng 5 2018
\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 1\)
\(\Rightarrow\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}< 1\left(đpcm\right)\)
6 tháng 5 2018
ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{100}}\)
\(A=1-\frac{1}{2^{100}}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
NT
1
9 tháng 5 2018
Ta có:3.A=1+1/3+1/3^2+...+1/3^97 +1/3^98
=>3.A - A=(1+1/3+1/3^2+...+1/3^98 + 1/3^99)-(1/3+1/3^2 +1/3^3+...+1/3^98+1/3^99)
=>2.A=1-1/3^99
=>A=1/2 -1/3^99.1/2 <1/2
Vậy ... T I C K cho mình với nha
MH
1
NT
3
Ta có : \(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)
\(2A=2+\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+...+\frac{2}{2^{2017}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{2}{2^{2016}}\right)\)
\(A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2016}}-\frac{1}{2^{2017}}\)
\(A=2-\frac{1}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)
Vậy A < 1
Q
27 tháng 4 2017
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)
\(A=2-\frac{1}{2^{2017}}\left(đpcm\right)\)
VN
1
6 tháng 5 2018
\(A=\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{96^2}+\frac{1}{98^2}\)
\(A< \frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}+\frac{1}{97.99}\)
\(A< \frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}+\frac{1}{99}\)
\(A< 1-\frac{1}{99}\)
\(A< \frac{98}{99}\)
A= \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+ \(\frac{1}{2^3}\)+...+ \(\frac{1}{2^{99}}\)+ \(\frac{1}{2^{100}}\).
2A= 1+ \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+...+ \(\frac{1}{2^{100}}\)+ \(\frac{1}{2^{101}}\).
2A- A=( 1+ \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+...+ \(\frac{1}{2^{100}}\)+ \(\frac{1}{2^{101}}\))-( \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+ \(\frac{1}{2^3}\)+...+ \(\frac{1}{2^{99}}\)+ \(\frac{1}{2^{100}}\)).
A= 1- \(\frac{1}{2^{100}}\)< 1.
=> A< 1.
Vậy A< 1.
Ta có
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow2A=\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+...+\frac{2}{2^{100}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
\(\Rightarrow A< 1\)
Vậy A<1 (đpcm)