\(A=1+3+3^2+3^3+...+3^{102}+3^{103}\)

\(\Rightarrow A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{102}+3^{103}\right)\)

\(\Rightarrow A=\left(1+3\right)+3^2\left(1+3\right)+...+3^{102}\left(1+3\right)\)

\(\Rightarrow A=\left(1+3\right)\left(1+3^2+...+3^{102}\right)\)

\(\Rightarrow A=4\left(1+3^2+...+3^{102}\right)⋮4\)

\(A=1+3^2+3^4+...+3^{102}\)

\(9A=3^2+3^4+...+3^{102}+3^{104}\)

\(\Rightarrow9A-A=3^{104}-1\)

\(\Rightarrow8A=3^{104}-1\)

\(\Rightarrow A=\dfrac{3^{104}-1}{8}\)

bài 3 : tính nhanh

a) (42*43+43*57+43)-360:4

ngoặc vuông43.(42+57+1) ngoặc vuông -90

43.100-90

4300-90

4210

b.

a) (42*43 + 43*57 + 43) - 360:4

= 43*(42+57+1) - 90

= 43*100 - 90

= 4300 - 90

= 3410

a) (42*43+43*57+43)-360:4

= 43. (42+ 57 +1) - 90

= 43 .100 - 90

= 4300 - 90 =4210

b) hình như đề sai 

c) 58*42+32*8+5+16

=58. 42 + 2.16 . 8 + 5+16

= 58 . 21.2 + 16 . (2.8 + 5)

= 58 . 21 . 2 + 16 . 21

= 21 . (58 .2 + 16)

= 21 . 132

= 2771

*HỌC TỐT*

TK NHÉ, B   ÊU ^^

Dịch ra là: Ta có: 3A = 3. (1 + 3 + 32 + 33 + ... + 399 + 3100) (1 + 3 + 32 + 33 + ... + 399 + 3100) 3A = 3 + 32 + 33 + ... + 3100 + 31013 + 32 + 33 + ... + 3100 + 3101 Suy ra: 3A - A = (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) ⇒⇒ A = 3101−123101−12 Vậy A = 3101−12

Mà đoạn 2A sai nhé bạn, sửa lại:

2A = 3101−13101−1 2A=-10001

A=-10001/2

A=-5000,5

Vậy A=-5000,5

\(12+3\cdot3^5:4-3\)

\(=4\cdot3+3^6\cdot\dfrac{1}{4}-3\)

\(=3\cdot\left(4+3^5\cdot\dfrac{1}{4}-1\right)\)

\(=3\cdot\left(3+3^5\cdot\dfrac{1}{4}\right)\)

\(=3\cdot3\cdot\left(1+3^4\cdot\dfrac{1}{4}\right)\)

\(=9\cdot\dfrac{85}{4}\)

\(=\dfrac{785}{4}\)

_________________

\(90^3-88\cdot3\cdot3^9\cdot3\)

\(=9^3\cdot10^3-88\cdot3^{11}\)

\(=\left(3^2\right)^3\cdot10^3-88\cdot3^{11}\)

\(=3^6\cdot1000-88\cdot3^{11}\)

\(=3^6\cdot\left(1000-88\cdot3^5\right)\)

\(=3^6\cdot-20384\)

\(=-14859936\)

________________

\(10^2+10^3+10^4+10^5+10^6\)

\(=\left(10^2+10^3\right)+\left(10^4+10^5\right)+10^6\)

\(=10^2\cdot\left(1+10\right)+10^4\cdot\left(1+10\right)+10^6\)

\(=11\cdot\left(10^2+10^4\right)+10^6\)

\(=11\cdot\left(100+10000\right)+10^6\)

\(=11\cdot10100+1000000\)

\(=111100+1000000\)

\(=1111100\)

_______________

\(7-3:2\cdot7^{10}\)

\(=7\cdot\left(1-\dfrac{3}{2}\cdot7^9\right)\)

\(=7\cdot-60530409,5\)

\(=-423712866,5\)

xin lỗi bài trên của mình làm sai

Ta có: 3A = 3.(1+3+3²+3³+...+3⁹⁹+3¹⁰⁰⁾ 

3A = 3+3²+3³+...+3¹⁰⁰+3¹⁰¹

Suy ra: 3A – A = (3+3²+3³+...+3¹⁰⁰+3¹⁰¹)−(1+3+3²+3³+...+3⁹⁹+3¹⁰⁰)

2A = 3¹⁰¹−1

⇒ A = 3¹⁰¹−1

             2               

Vậy A = 3¹⁰¹−1

                 2           

em den lam

\(D=3^{100}+3^{101}+...+3^{149}+3^{150}\)

nên \(3D=3^{101}+3^{102}+...+3^{150}+3^{151}\)

\(\Leftrightarrow2\cdot D=3^{151}-3^{100}\)

hay \(D=\dfrac{3^{151}-3^{100}}{2}\)

\(3D=3^{101}+3^{102}+3^{103}+...+3^{150}+3^{151}\\ 3D-D=3^{151}-3^{100}\\ 2D=3^{151}-3^{100}\\ D=\dfrac{3^{151}-3^{100}}{2}\)

2884 nha

103 sản phẩm

1878 nha

Lời giải:
$A=(1+3+3^2+3^3)+(3^4+3^5+3^6+3^7)+....+(3^{56}+3^{57}+3^{58}+3^{59})$

$=(1+3+3^2+3^3)+3^4(1+3+3^2+3^3)+...+3^{56}(1+3+3^2+3^3)$

$=(1+3+3^2+3^3)(1+3^4+...+3^{56})$

$=40.(1+3^4+...+3^{56})\vdots 10$

Do đó chữ số tận cùng của $A$ là $0$