\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{2017.2017}\)

Ta có :

\(\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3.3}< \frac{1}{2.3}\)

\(\frac{1}{4.4}< \frac{1}{3.4}\)

........

\(\frac{1}{2017.2017}< \frac{1}{2016.2017}\)

=> \(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{2017.2017}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2016.2017}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2016}-\frac{1}{2017}\)

\(=1-\frac{1}{2017}< 1\)

=> A < 1 

\(a=\frac{1}{2.2}+\frac{1}{3.3}+........+\frac{1}{2017.2017}\)

\(a< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2016.2017}\)

\(a< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2016}-\frac{1}{2017}\)

\(a< 1-\frac{1}{2017}\)

Do \(a< 1-\frac{1}{2017}\)

  \(\Rightarrow a< 1\)

A = 1/1×2 + 1/2×3 + 1/3×4 + .. + 1/99×100

A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

A = 1 - 1/100 < 1

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(A=1\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

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

=>  ĐPCM

Ta có:

A = 1/1 x 2 + 1/2 x 3 + 1/3 x 4 + ..... + 1/99 x 100 

A = 1- 1/2 + 1/2 - 1 /3 + 1/3 - 1/4 + ..... + 1/99 - 1/100 

A = 1 - 1/100 < 1 

nha bn 

chúc bn học giỏi

A là số dương, B là số âm => A>B

A có 50 thừa số âm

=> A > 0

b) CÓ 49 thừa số âm 

=> B < 0 

A có 50 thừa số âm

=> A > 0

B có 49 thừa số âm

=> B < 0

tick nha

A<0

B>0

bang 2 cu bam may tinh la ra!

4A=1+1/4+1/4²+......+1/4⁹⁸

4A - A = ( 1+1/4+1/4²+..........+1/4⁹⁸) - ( 1/4+1/4²+1/4³+.......+1/4⁹⁹)

3A= 1-1/4⁹⁹

A= 1/3 - 1/4⁹⁹ : 3

Mà 1/4⁹⁹ : 3 > 0 => 1/3 - 1/4⁹⁹ : 3 < 1/3

                          Hay A < 1/3

a/ Rút gọn:

\(A=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{99}}.\)

=> \(4A=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{98}}\)

=> \(4A=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{98}}+\frac{1}{4^{99}}\right)-\frac{1}{4^{99}}\)

<=> \(4A=1+A-\frac{1}{4^{99}}\)

=> \(3A=1-\frac{1}{4^{99}}\)

=> \(A=\frac{1}{3}-\frac{1}{3.4^{99}}\)

b/ Ta có: \(A=\frac{1}{3}-\frac{1}{3.4^{99}}< \frac{1}{3}\)

Ta có:

A=1/4+1/4²+…+1/4⁹⁹

4A=4.(1/4+1/4²+…+1/4⁹⁹)

4A=1+1/4+…+1/4⁹⁸

4A-A=(1+1/4+…+1/4⁹⁸)- (1/4+1/4²+…+1/4⁹⁹)

3A=1-1/4⁹⁹

A=1/3-1/(4⁹⁹.3)

b)Vì A=1/3-1/(4⁹⁹​.3) nên A<1/3 (do 1/4⁹⁹​.3>0)

Ta có: \(S=1+3+3^2+...+3^{20}\)

\(\Rightarrow3S=3+3^2+3^3+...+3^{21}\)

\(\Rightarrow3S-S=\left(3+3^2+3^3+...+3^{21}\right)-\left(1+3+3^2+...+3^{20}\right)\)

\(\Rightarrow2S=3^{21}-1\)

\(\Rightarrow S=\left(3^{21}-1\right).\frac{1}{2}\)

\(\Rightarrow S=3^{21}.\frac{1}{2}-\frac{1}{2}\)

Vì \(3^{21}.\frac{1}{2}-\frac{1}{2}< 3^{21}.\frac{1}{2}\) nên \(A< \frac{1}{2}.3^{21}\)

Vậy \(A< \frac{1}{2}.3^{21}\)