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``.`

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

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

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

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

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

\(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

Giải:

A=1/2²+1/3²+1/4²+...+1/9²

Ta có:

1/2²<1/1.2

1/3²<1/2.3

1/4²<1/3.4

...

1/9²<1/8.9

⇒A<1/1.2+1/2.3+1/3.4+...+1/8.9

A<1/1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9

A<1/1-1/9

A<8/9

Ta có:

1/2²>1/2.3

1/3²>1/3.4

1/4²>1/4.5

...

1/9²>1/9.10

⇒A>1/2.3+1/3.4+1/4.5+...+1/9.10

A>1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10

A>1/2-1/10

A>2/5

Vậy 2/5<A<8/9 (đpcm)

Chúc bạn học tốt!

bít kq nhưng ko thích giải

cậu ko giúp cậu ấy thì thôi đừng bảo như thế

1/5+1/9+1/10+1/41+1/42 < 1/2

Bạn phải giải đầy đủ ra thì mình mới k cho

1/5+1/9+1/10+1/41+1/42=5932/12915suy ra 5932/12915<1/2

S<1/2 (tính ra S ròi ss thui)

Bài 1

a) S = 1 + 2 + 2² + 2³ + ... + 2²⁰²³

2S = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²⁴

S = 2S - S = (2 + 2² + 2³ + ... + 2²⁰²⁴) - (1 + 2 + 2² + 2³)

= 2²⁰²⁴ - 1

b) B = 2²⁰²⁴

B - 1 = 2²⁰²⁴ - 1 = S

B = S + 1

Vậy B > S

a,

\(S=1+2+2^2+...+2^{2023}\)

\(2S=2+2^2+2^3+...+2^{2024}\)

\(\Rightarrow S=2^{2024}-1\)

b.

Do \(2^{2024}-1< 2^{2024}\)

\(\Rightarrow S< B\)

2.

\(H=3+3^2+...+3^{2022}\)

\(\Rightarrow3H=3^2+3^3+...+3^{2023}\)

\(\Rightarrow3H-H=3^{2023}-3\)

\(\Rightarrow2H=3^{2023}-3\)

\(\Rightarrow H=\dfrac{3^{2023}-3}{2}\)

Bài 2

H = 3 + 3² + 3³ + ... + 3²⁰²²

⇒ 3H = 3² + 3³ + 3⁴ + ... + 3²⁰²³

⇒2H = 3H - H

= (3² + 3³ + 3⁴ + ... + 3²⁰²³) - (3 + 3² + 3³ + ... + 3²⁰²²)

= 3²⁰²³ - 3

⇒ H = (3²⁰²³ - 3) : 2

Ta thấy:

\(2^2=2.2>1.2\Rightarrow\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(3^2=3.3>2.3\Rightarrow\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

.................

\(9^2=9.9>8.9\Rightarrow\dfrac{1}{9^2}< \dfrac{1}{8.9}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)

\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}=1-\dfrac{1}{9}=\dfrac{8}{9}\)

=> Đpcm

Ta thấy:

22=2.2>1.2⇒122<11.222=2.2>1.2⇒122<11.2

32=3.3>2.3⇒132<12.332=3.3>2.3⇒132<12.3

.................

92=9.9>8.9⇒192<18.992=9.9>8.9⇒192<18.9

⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9

⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89

=> ...(tự viết)

Ta thấy:

22=2.2>1.2⇒122<11.222=2.2>1.2⇒122<11.2

32=3.3>2.3⇒132<12.332=3.3>2.3⇒132<12.3

.................

92=9.9>8.9⇒192<18.992=9.9>8.9⇒192<18.9

⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9

⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89

=> 11111111111111111111110101010110000

HACK

cho sửa lại là:

Ta thấy:

\(⇒ \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... < \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

\(⇔\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2} <1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4} ...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)

⇒ {{{tự làm nhé}}}