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

\(S=1+2+2^2+2^3+...+2^9\) 

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``.`

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

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

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

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

\(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

\(S=1+2+2^2+2^3+2^4+...+2^{2011}\)

\(\Rightarrow S=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{2009}\left(1+2+2^2\right)\)

\(\Rightarrow S=7+2^3.7+...+2^{2009}.7\)

\(\Rightarrow S=7\left(1+2^3+...+2^{2009}\right)⋮7\)

\(\Rightarrow dpcm\)

vì tổng của S chia hết cho 3 nên S chia hết cho 3. có thế cũng hỏi =))

Chúc bạn an toàn

có phép trừ ko

nếu ko có thì tổng đó lớn hơn 251

rõ ràng mà

s=[1+2]+[2+2 mũ 2]+...+[2 mũ 6+2 mũ 7]

s=1 nhân [1+2]+2 nhân [1+2]+...+2 mũ 6 nhân [1+2]

s=[1+2] nhân[1+2+...+2 mũ 6

s=3 nhân [1+2+...+2 mũ 6]

=> s chia hết cho 3

\(S=\left(1+2\right)+...+2^6\left(1+2\right)=3\left(1+...+2^6\right)⋮3\)

\(S=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{95}+2^{96}\right)\\ S=\left(1+2\right)\left(2+2^3+...+2^{95}\right)\\ S=3\left(2+2^3+...+2^{95}\right)⋮3\left(1\right)\\ S=\left(2+2^2\right)+2^3\left(1+2^2+...+2^{93}\right)\\ S=8+8\left(1+2^2+...+2^{93}\right)⋮8\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow S⋮24\)

Sửa đề: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)

Ta có: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)

\(=\dfrac{1}{20}+\left(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{30}\right)+\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)\)

\(\Leftrightarrow S>\dfrac{1}{20}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{4}\)

\(\Leftrightarrow S>\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)(đpcm)

2/3+3/4+...=2+1/2