1+2+2^2 + ...+2^19> tick cho mk mk trình bày cụ thể cho

2 x a = 2+2^2+...+2^19+2^20
khi đó 2xa-a=[ 2+2^2+...+2^19+2^20] - [1+2+...+2^19]
a=2^20-1
nhìn vào bài b=2^20
so sánh a và b ta thấy a<b

vậy a<b

tích nha chuẩn luôn 100%

\(A=1+2+2^2+...+2^{101}\)

\(2A=2+2^2+...+2^{102}\)

\(2A=\left(2+2^2+...+2^{102}\right)-\left(1+2+2^2+...+2^{101}\right)\)

\(A=2^{102}-1\)

\(B=5.2^{100}>2^{102}\)

Mà \(2^{102}>2^{102}-1\)

Nên B>A

S=1+2+2^2+2^3+...+2^20

2.S=2+2^2+2^3+...+2^20+2^21

2.S-S=S=(2+2^2+2^3+....+2^21)-(1+2+2^2+...+2^20)

S=2^21-1

bây giờ so sánh 2^21-1 với 5.2^19

mà 2^21-1=2^19.2^2-1 hay 2^19 .4 -1 <2^19.5

=>S<2^19.5

\(S=1+2+2^2+2^3+\cdots+2^{20}\)

=> \(2S=2+2^2+2^3+2^4+\cdots+2^{21}\)

=> \(2S-S=\left(2+2^2+2^3+2^4+\cdots+2^{21}\right)-\left(1+2+2^2+2^3+\ldots+2^{20}\right)\)

=>\(S=2^{21}-1\)

Mà \(2^{21}-1=2^{19}.2^2-1\) hay \(2^{19}.4-1<2^{19}.5\)

=>\(S<2^{19}.5\)

Ta có: \(S=1+2+2^2+\cdots+2^{20}\)

=>\(2S=2+2^2+2^3+\cdots+2^{21}\)

=>\(2S-S=2+2^2+\cdots+2^{21}-1-2-\cdots-2^{20}\)

=>\(S=2^{21}-1\)

\(2^{21}-1-5\cdot2^{19}=2^{19}\left(2^2-5\right)-1=-2^{19}-1<0\)

=>\(S-5\cdot2^{19}<0\)

=>\(S<5\cdot2^{19}\)

ta có: \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}+\frac{1}{20}\)( Có 10 phân số 1/20)

                                                                                            \(=\frac{10}{20}=\frac{1}{2}\)

\(\Rightarrow A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}>\frac{1}{2}\)

Chúc bn học tốt !!!!

cảm ơn

Áp dụng tính chất (a - b)(a + b) = a² + ab - ab - b² = a² - b²

Ta có : \(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{19}{\left(9.10\right)^2}\)

\(=\frac{1}{1.2}.\frac{3}{1.2}+\frac{1}{2.3}.\frac{5}{2.3}+...+\frac{1}{9.10}.\frac{19}{9.10}\)

\(=\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)\left(\frac{1}{2}+\frac{1}{3}\right)+...+\left(\frac{1}{9}-\frac{1}{10}\right)\left(\frac{1}{9}+\frac{1}{10}\right)\)

\(=1^2-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2-\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{9}\right)^2-\left(\frac{1}{10}\right)^2=1^2-\left(\frac{1}{10}\right)^2=1-\frac{1}{100}=\frac{99}{100}< 1\)

\(2A=2+2^2+2^3+2^4+...+2^{50}\)

\(A=2A-A=2^{50}-1\)

Ta có \(2^{50}=2.2^{49}=2.\left(2^7\right)^7=2.128^7\)

\(5^{19}< 5^{21}=\left(5^3\right)^7=125^7\)

\(\Rightarrow125^7< 128^7< 2.128^7\Rightarrow5^{19}< 2^{50}\Rightarrow5^{19}-1< 2^{50}-1=A\)

1/11 +  1/12+ 1/13 +...+ 1/19 + 1/20 > 1/20 + 1/20 + 1/20 + ... + 1/20 

= 10/20 = 1/2 

Vậy A > 1/2.