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1 tháng 5 2018

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.

1 tháng 5 2018

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)

A=12.34.56...99100

⇒A<23.45.67...100101

⇒A2<23.45.67...100101.12.34.56...99100

⇒A2<1101<1100=1102

A=12.34.56...99100

⇒A<23.45.67...100101

⇒A2<23.45.67...100101.12.34.56...99100

⇒A2<1101<1100=1102

29 tháng 11 2021

\(\Rightarrow3B=3^2+3^3+3^4+...+3^{101}\\ \Rightarrow3B-B=3^2+3^3+...+3^{101}-3-3^2-3^3-...-3^{100}\\ \Rightarrow2B=3^{101}-3\\ \Rightarrow B=\dfrac{3^{101}-3}{2}\)

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

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: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=1-\frac{1}{2^{100}}\)

\(A=1-\frac{1}{2^{100}}< 1\)

\(\Rightarrow A< 1\left(đpcm\right)\)

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

22 tháng 4 2016

Tinh 2A, roi lay 2A-A se chung to dc

27 tháng 4 2017

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 

27 tháng 4 2017

\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)

\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}\)

\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)

\(A=2-\frac{1}{2^{2017}}\left(đpcm\right)\)

6 tháng 5 2018

\(A=\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{96^2}+\frac{1}{98^2}\)

\(A< \frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}+\frac{1}{97.99}\)

\(A< \frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}+\frac{1}{99}\)

\(A< 1-\frac{1}{99}\)

\(A< \frac{98}{99}\)