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

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

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

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

\(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

\(S=1+2+2^2+2^3+...+2^9\) 

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``.`

ta có 1/3=10/30

1/21+1/22+...+1/30 có 10 p/số

mà 1/21>1/30

1/22>1/30

....

1/29>1/30

1/30=1/30

=>1/21+..1/30>1/30+....1/30 có 10 phân số 

=>1/21+...1/30>1/3

Ta có: \(\frac{1}{21}< \frac{1}{30}\)

\(\frac{1}{22}< \frac{1}{30}\)

......

\(\frac{1}{29}< \frac{1}{30}\)

\(\Rightarrow S< \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\)(có 10 p/s)
\(\Rightarrow S< \frac{1}{30}.10=\frac{10}{30}=\frac{1}{3}\)

Vậy S < 1/3

ta co 1/21+1/22+1/23>3/30

1/24+1/25+1/26>3/30

1/27+1/28+1/29>3/30

==>S>3/30+3/30+3/30+1/30

S>10/30 hay S>1/3

Số số hạng của tổng A là : \(\dfrac{30-21}{1}+1=10\left(sh\right)\)

`=>A=\underbrace{1/21+1/22+...+1/30}_{10sh}>\underbrace{1/30+1/30+1/30+...+1/30}_{10sh}`

`=>A>(1)/(30).10`

`=>A>10/30`

`=>A>1/3`

`=>đpcm`

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

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

1/

Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.

Số số hạng: $(101-1):4+1=26$

$A=(101+1)\times 26:2=1326$

2/

$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$

$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$

$=(1+2+2^2)(1+2^3+2^6+2^9)$

$=7(1+2^3+2^6+2^9)\vdots 7$

3/
$C=1+2+2^2+2^3+...+2^{99}$

$2C=2+2^2+2^3+2^4+...+2^{100}$

$\Rightarrow 2C-C=2^{100}-1$

$\Rightarrow C=2^{100}-1$

a. bằng77

b. bằng 13

lớp 2 đã học cái này rồi à

a. bằng77

b. bằng 13

S > 1/3

ta thấy \(\frac{1}{20}\)<\(\frac{1}{3}\)

thì \(\frac{1}{20}\)+...+\(\frac{1}{29}\)<\(\frac{1}{20}\)+...+\(\frac{1}{20}\)<\(\frac{1}{3}\)

vậy \(\frac{1}{20}\)+...+\(\frac{1}{29}\)<\(\frac{1}{3}\)

bít kq nhưng ko thích giải

cậu ko giúp cậu ấy thì thôi đừng bảo như thế