Đặt \(D=3-3^2+3^3-3^4+...+3^9-3^{10}+3^{11}\)

=> \(3D=3^2-3^3+3^4-3^5+...+3^{10}-3^{11}+3^{12}\)

Cộng vế 2 BT trên ta được:

\(D+3D=\left(3-3^2+...+3^{11}\right)+\left(3^2-3^3+...+3^{12}\right)\)

\(\Leftrightarrow4D=3^{12}+3\)

\(\Rightarrow D=\frac{3^{12}+3}{4}\)

\(3^6:3^2+2^3.2^2-3^3.3\)

\(=3^4+2^5-3^4\)

\(=3^4-3^4+2^5\)

\(=0+2^5=2^5\)

\(3^6:3^2+2^3.2^2-3^3.3\\ =3^4+2-3^4\\ =\left(3^4-3^4\right)+2\\ =0+2\\ =2.\)

Sửa hộ mk thành \(2^5\) bắt đầu từ dòng thứ 2 nhé.

a, 2¹.5².17 = 2.25.17 = 50.17 = 850 

b, 2² + 2³ + 2⁴ = 4 + 8 + 16 = 28

c, 2⁵.3 + 2⁴:8 + 50: 5²

= 32.3 + 16:8 + 50:25

=96 + 2 + 2

= 100

d, 11² - 10² - 3²

= 121 - 100 - 9

= 21 - 9

= 12

e, 1³ + 2³ + 3³ + 4³ + 5³

= ( 1+ 2+3+4+5)²

= 15²

= 225

Giúp mình giải bài này nha

G=1-3+3²-3³+3⁴-...-3⁹⁹+3¹⁰⁰

3G=3-3²+3³-3⁴+3⁵-....-3¹⁰⁰+3¹⁰¹

3G+G=(3-3²+3³-3⁴+3⁵-....-3¹⁰⁰+3¹⁰¹)+(1-3+3²-3³+3⁴-...-3⁹⁹+3¹⁰⁰)

4G = 3¹⁰¹+1

G=\(\frac{3^{101}+1}{4}\)

\(\Rightarrow3S=3^2+3^3+3^4+...+3^{2023}\)

trừ vế với vế ta được :

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

\(\Rightarrow2S=3^{2023}-3\)

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

\(A=3^1+3^2+3^3+3^4+...+3^{199}\)

\(3A=3^2+3^3+3^4+3^5+...+3^{200}\)

\(3A-A=\left(3^2+3^3+3^4+...+3^{200}\right)-\left(3^1+3^2+3^3+...+3^{199}\right)\)

\(2A=3^{200}-3^1\)

\(A=\frac{3^{200}-3}{2}\)

=))

Đặt \(A=3^1+3^2+3^3+...+3^{199}\)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{200}\)

Lấy 3A trừ A theo vế ta có : 

\(3A-A=\left(3^2+3^3+3^4+..+3^{200}\right)-\left(3^1+3^2+3^3+..+3^{199}\right)\)

\(2A=3^{200}-1\)

\(A=\frac{3^{200}-1}{2}\)

Vậy \(3^1+3^2+3^3+..+3^{199}=\frac{3^{200}-1}{2}\)

xích ma 3x chạy từ 1 tới 199 kết quả là \(^{\text{1,328069944 nhân}10^{95}}\)

S=3+3^2+3^3+...+3^2022

3S=3.(3+3^2+3^3+...+3^2022)

3S=3^2+3^3+3^4+...+3^2023

⇒3S-S=(3^2+3^3+3^4+...+3^2023)-(3+3^2+3^3+...+3^2022)

⇒2S=3^2023-3

⇒S=3^2023-3 / 2

S=3+3^2+3^3+...+3^2022

=>3S=3^2+3^3+3^4+...+3^2023

=>3S-S=(3^2+3^3+3^4+...+3^2023)-(3+3^2+3^3+...+3^2022)

=>2S=3^2023-3

=>S=\(\dfrac{3^{2023}-3}{2}\)

Vậy S=​\(\dfrac{3^{2023}-3}{2}\)​

CHÚC BẠN HỌC TỐT NHA!