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

3 mũ 11:52 là số nào vậy bạn

\(Q=1+3+3^2+3^3+3^4+...+3^{11}\)

\(3Q=3+3^2+3^3+3^4+3^5+...+3^{12}\)

\(3Q-Q=\left(3+3^2+3^3+3^4+3^5+...+3^{12}\right)-\left(1+3+3^2+3^3+3^4+...+3^{11}\right)\)

\(2Q=3^{12}-1\)

\(Q=\frac{3^{12}-1}{2}\)

Ta có: \(S=3+3^2+3^3+\cdots+3^{2024}\)

\(=\left(3+3^2\right)+\left(3^3+3^4+3^5\right)+\left(3^6+3^7+3^8\right)+\cdots+\left(3^{2022}+3^{2023}+3^{2024}\right)\)

\(=12+3^3\left(1+3+3^2\right)+3^6\left(1+3+3^2\right)+\cdots+3^{2022}\left(1+3+3^2\right)\)

\(=12+13\left(3^3+3^6+\cdots+3^{2022}\right)\)

=>S không chia hết cho 13

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

trừ vế với vế ta được :

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

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

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

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=​\(\dfrac{3^{2023}-3}{2}\)​