Theo đề bài ra, ta có :

`A=1+32+34+36+....+32008`

\(\Rightarrow\) `9A = 3^2 + 3^4 + 3^6 + 3^8 + ... + 3^2010`

`9A - A=(32+34+36+38+....+ 32010)-(1+32+34+36+....+ 32008)`

\(\Rightarrow\) `8A=(-1)+32010`

\(\Rightarrow\) `8A-32010=(-1)`

@Nae

Theo đề bài ra, ta có :

A=1+3^2+3^4+3^6+....+3^2008

 9A = 3^2 + 3^4 + 3^6 + 3^8 + ... + 3^2010

9A - A= (3^2+3^4+3^6+3^8+....+ 3^2010)- (1+3^2+3^4+3^6+....+ 3^2008)

8A = -1+3^2010

 8A - 3^2010 = (-1)

@Nae

B = 1 + 32 + 34 + … + 32018

32.B = 32.( 1 + 32 + 34 + … + 32018)

9B = 32 + 34 + 36 + … + 32020

9B – B = (32 + 34 + 36 + … + 32020) – (1 + 32 + 34 + … + 32018)

8B = 32020 – 1

B = (32020 – 1) : 8.

Vậy B = (32020 – 1) : 8.

tick cho mink nhé (●'◡'●)

B = (3²⁰²⁰ - 1) : 8

a=0,1,2,3,4,5,6,7,8,9

b=1

c=1000,2000,3000,4000,5000,6000,7000,8000,9000.

2.

Trong dãy số trên có tất cả số số hạng là: ( 40

-2):2+1=20(số) ​​ Tổng của các số trên là:(40+2)

​

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a) \(A=2+2^2+2^3+...+2^{2017}\)

\(2A=2^2+2^3+2^4+...+2^{2018}\)

\(2A-A=\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\)

\(A=2^{2018}-2\)

b) \(C=1+3^2+3^4+...+3^{2018}\)

\(3^2\cdot C=3^2+3^4+3^6+...+3^{2020}\)

\(9C-C=\left(3^2+3^4+3^6+...+3^{2020}\right)-\left(1+3^2+3^4+...+3^{2018}\right)\)

\(8C=3^{2020}-1\)

\(\Rightarrow C=\dfrac{3^{2020}-1}{8}\)

\(Toru\)

\(25\cdot16-12\cdot55\)

\(=400-660\)

\(=-260\)

\(---\)

\(930:15-320:5+196\)

\(=62-64+196\)

\(=-2+196\)

\(=194\)

\(---\)

\(4\cdot5^2-81:3^2\)

\(=4\cdot25-81:9\)

\(=100-9\)

\(=91\)

\(---\)

\(7^6:7^4+3^4:3^2-3^7:3^6\)

\(=7^2+3^2-3\)

\(=49+9-3\)

\(=58-3\)

\(=55\)

#\(Toru\)

a) 576 + 678 + 780 - 475 - 577 - 679

= 780 - 475 + ( 576 - 577) - ( 678 - 679)

= 305 - 1 - 1

= 303

b) (126 + 32)\(\times\)(18 - 16 - 2)

= (126 + 32) \(\times\)0

= 0

c) 36 * 17 * 12 * 34 + 6 * 30

= 36 * 17 * 6 * 2 * 34 + 6 * 30

= 6 * (36 + 17 + 2 + 34 + 30)

= 6 * 119

= 714

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

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Bài 1:

a. $2^{29}< 5^{29}< 5^{39}$

$\Rightarrow A< B$

b.

$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$

$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$

$=(1+3)(3+3^3+3^5+...+3^{2009})$

$=4(3+3^3+3^5+...+3^{2009})\vdots 4$

Mặt khác:

$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$

$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$

Bài 1:
c.

$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$

$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$

$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$

$\Rightarrow A=\frac{3^{101}+1}{4}$

Bài 2:

a. $7\vdots n+1$

$\Rightarrow n+1\in \left\{1; -1; 7; -7\right\}$

$\Rightarrow n\in \left\{0; -2; 6; -8\right\}$

b.

$2n+5\vdots n+1$
$\Rightarrow 2(n+1)+3\vdots n+1$

$\Rightarrow 3\vdots n+1$

$\Rightarrow n+1\in \left\{1; -1; 3; -3\right\}$

$\Rightarrow n\in \left\{0; -2; 2; -4\right\}$

\(a,=\left(576-577\right)+\left(678-679\right)+\left(780-475\right)=-1-1+305=303\\ b,=158\times0=0\\ c,=12\left(36\times17\times34+15\right)=12\times20823=249876\)

Ta có :

x = 31 + 32 - 33 + 34 - 35 + 36 - 37 + 38 - 39 + 40

x = 31 - 32 + 33 - 34 + 35 - 36 + 37 - 38 + 39 + 40

x = 31 - ( 33 - 32 ) - ( 35 - 34 ) - ( 37 - 36 ) - ( 39 - 38 ) + 40

x = 31 - 1 - 1 - 1 - 1 + 40

x = 67

Vậy giá trị của biểu thức trên là 67

giá trị của biểu thức là 67