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!

a/

\(3S=3+3^2+3^3+3^4+...+3^{120}\)

\(2S=3S-S=3^{120}-1\Rightarrow S=\frac{3^{120}-1}{2}\)

b/ \(S=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)

\(S=13+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)

\(S=13+3^3.13+...+3^{117}.13=13\left(1+3^3+...+3^{117}\right)\) chia hết cho 13

c/

\(S=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)

\(S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{116}\left(1+3+3^2+3^3\right)\)

\(S=40+3^4.40+...+3^{116}.40=40\left(1+3^4+...+3^{116}\right)\) chia hết cho 40

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

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

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

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

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

=))

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 \(^{\text{1,328069944 nhân}10^{95}}\)

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

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

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

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

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

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

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

a)3¹x3²x3³x........x3¹⁰⁰

=3^{1+2+3+4+...+100}

=3^{(100+1)x(100-1+1):2}

=3^101x100:2

=3⁵⁰⁵⁰

Bài b mình không biết làm

thank nha

\(A=\)\(-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...+\frac{1}{3^{50}}-\frac{1}{3^{51}}\)

\(3A=-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{49}}-\frac{1}{3^{50}}\)

\(4A=-1-\frac{1}{3^{51}}\)

\(A=\frac{-1-\frac{1}{3^{51}}}{4}\)

k cho mik nha