Có : \(S=1+2+2^2+2^3+....+2^{99}\)

\(\Rightarrow2S=2+2^2+2^3+....+2^{100}\)

\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+....+2^{99}\right)\)

\(\Rightarrow S=2^{100}-1< 2^{100}\)

Vậy \(S< 2^{100}\)

 S=1+2+2²+2³+....+2⁹⁹

⇒2S=2+2²+2³+....+2¹⁰⁰

⇒2S−S=2¹⁰⁰-1

S=2¹⁰⁰-1

vì 2¹⁰⁰ -1<2¹⁰⁰

⇒S<2¹⁰⁰

\(M=\frac{1}{201}+\frac{1}{202}+...+\frac{1}{299}+\frac{1}{300}\)

\(\Rightarrow\)Có 100 phân số

Ta có: \(\frac{1}{201}>\frac{1}{300}\)

          \(\frac{1}{202}>\frac{1}{300}\)

             ...................

            \(\frac{1}{299}>\frac{1}{300}\)

            \(\frac{1}{300}=\frac{1}{300}\)

\(\Rightarrow M>\left(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)=\frac{100}{300}=\frac{1}{3}\)

Vậy....

\(S=1+2+2^2+2^3+...+2^9\) 

Đặt \(2S=2+2^2+2^3+2^4+...+2^{10}\) 

\(2S-S=2^{10}-1\) hay \(S=2^{10}-1< 2^{10}\)

\(\Rightarrow\) \(2^{10}=2^2.2^8< 5.2^8\) 

Vậy \(S< 5.2^8\)

\(#Tuyết\)

2S=2+2^2+...+2^10

=>S=2^10-1=1023

5*2^8=256*5=1280

=>S<5*2^8

`@` `\text {Answer}`

`\downarrow`

`S = 1 + 2 + 2^2 + 2^3 + ... + 2^9`

`=> 2S = 2 + 2^2 + 2^3 + ... + 2^10`

`=> 2S - S = (2+2^2 + 2^3 + ... + 2^10) - (1 + 2 + 2^2 + 2^3+...+2^9)`

`=> S = 2^10 - 1`

Mà `2^10 - 1 < 2^10`

`=> S < 2^10 (1)`

Ta có:

 `2^10 = 2^7*8`

Mà `5*2^8 = 5* 2 * 2^7 = 10* 2^7`

Vì `10 > 8 => 2^7 * 8 < 2^7  * 10 (2)`

Từ `(1)` và `(2)`

`=> S < 5 * 2^7``.`

Vì: \(\frac{3}{21}=\frac{3}{21}\)

\(\frac{3}{22}\) < \(\frac{3}{21}\)

\(\frac{3}{23}\) < \(\frac{3}{21}\)

\(\frac{3}{24}\)<\(\frac{3}{21}\)

\(\frac{3}{25}\)< \(\frac{3}{21}\)

.....

\(\frac{2}{29}\)<\(\frac{3}{21}\)

\(\frac{2}{30}\)<\(\frac{3}{21}\)

Nên \(\frac{3}{21}+\frac{3}{22}+\frac{3}{23}+\frac{3}{24}+\frac{3}{25}+...+\frac{3}{29}+\frac{3}{30}\) < \(\frac{3}{21}.10\)

Ta có: \(\frac{3}{21}.10\) = \(\frac{10}{7}\)

Mà \(\frac{10}{7}\) < \(\frac{3}{2}\)

=>\(\frac{3}{21}+\frac{3}{22}+\frac{3}{23}+\frac{3}{24}+\frac{3}{25}+...+\frac{3}{29}+\frac{3}{30}\) < \(\frac{3}{2}\)

Vậy E < M

giups mình với

1+2+2²+2³+......2²⁰²²>5.2²²¹

2A=2*(1+2+2²+...+2²⁰²⁰)=2+2²+...+2²⁰²¹

^2A-A=(1+2+2²+...+2²⁰²¹)-(1+2+2²+...+2²⁰²⁰)

A=2²⁰²¹-1<2021

Giải:

A=1+2+2²+2³+...+2²⁰²⁰

2A=2+2²+2³+2⁴+...+2²⁰²¹

2A-A=(2+2²+2³+2⁴+...+2²⁰²¹)-(1+2+2²+2³+...+2²⁰²⁰)

A=2²⁰²¹-1

⇒A<2²⁰²¹

Chúc bạn học tốt!

$S=1+2+2^2+...+2^9$

$2S=2\left(1+2+2^2+...+2^{10}\right)$

$2S=2+2^2+2^3+...+2^9$

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

\(S=2^{10}-1=2^2.2^8-1=4.2^8-1

HT

\(S=1+2+2^2+...+2^9\)

\(S=\dfrac{2^{9+1}-1}{2-1}\)

\(S=2^{10}-1=1023\)

\(5.2^8=5.256=1280>1023\)

\(\Rightarrow S< 5.2^8\)

S < 5. 2^8

Là \(\left(\dfrac{1}{2}\right)^2\) hay \(\dfrac{1}{2^2}\) vậy bạn

Những cái sau tương tự

\(M=\frac{3}{1^22^2}+\frac{5}{2^23^2}+\frac{7}{3^24^2}+...+\frac{4019}{2009^22010^2}\)

\(M=\frac{2^2-1^2}{1^22^2}+\frac{3^2-2^2}{2^23^2}+\frac{4^2-3^2}{3^24^2}+...+\frac{2010^2-2009^2}{2009^22010^2}\)

\(M=\frac{2^2}{1^22^2}-\frac{1^2}{1^22^2}+\frac{3^2}{2^23^2}-\frac{2^2}{2^23^2}+\frac{4^2}{3^24^2}-\frac{3^2}{3^24^2}+...+\frac{2010^2}{2009^22010^2}-\frac{2009^2}{2009^22010^2}\)

\(M=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2009^2}-\frac{1}{2010^2}\)

\(M=1-\frac{1}{2010^2}< 1\)

Vậy \(M< 1\)

Chúc bạn học tốt ~