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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
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
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=\(\dfrac{3^{2023}-3}{2}\)
a)31x32x33x........x3100
=31+2+3+4+...+100
=3(100+1)x(100-1+1):2
=3101x100:2
=35050
Bài b mình không biết làm
Sửa đề: \(S=1-3+3^2-3^3+\ldots+3^{98}-3^{99}\)
=>3S=\(3-3^2+3^3-3^4+\ldots+3^{99}-3^{100}\)
=>3S+S=\(3-3^2+3^3-3^4+\cdots+3^{99}-3^{100}+1-3+3^2-3^3+\cdots+3^{98}-3^{99}\)
=>4S=\(-3^{100}+1\)
=>\(S=\frac{-3^{100}+1}{4}\)
S = 1-3 mũ 100/ 4
con ma đó
bieets vì sao ko
vì mìn ko hiểu mìn mới học lop4