cho S=32+33+...+32023 CMR tổng S chia hết cho 156
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(A=1+3+3^2+3^3+3^4+\cdot\cdot\cdot+3^{2023}+3^{2024}\)
\(=(1+3+3^2)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+\dots+(3^{2022}+3^{2023}+3^{2024})\\=13+3^3\cdot(1+3+3^2)+3^6\cdot(1+3+3^2)+\dots+3^{2022}\cdot(1+3+3^2)\\=13+3^3\cdot13+3^6\cdot13+\dots+3^{2022}\cdot13\\=13\cdot(1+3^3+3^6+\dots+3^{2022})\)
Vì \(13\cdot(1+3^3+3^6+\dots+3^{2022})\vdots13\)
nên \(A\vdots13\)
\(\Rightarrowđpcm\)
a: \(A=1+3+3^2+\cdots+3^{2022}\)
=>\(3A=3+3^2+3^3+\cdots+3^{2023}\)
=>\(3A-A=3+3^2+\cdots+3^{2023}-1-3^{}-\cdots-3^{2022}\)
=>\(2A=3^{2023}-1\)
=>\(2A-3^{2023}=-1\)
b: x+10⋮x-1
=>x-1+11⋮x-1
=>11⋮x-1
=>x-1∈{1;-1;11;-11}
=>x∈{2;0;12;-10}
\(S=1+3+3^2+3^3+...+3^8+3^9\)
\(=1+3+3^2\left(1+3\right)+...+3^8\left(1+3\right)\)
\(=4\left(1+3^2+...+3^8\right)⋮4\)
\(S=\left(1+3\right)+3^2\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+3^2+...+3^8\right)⋮4\)
Đây là toán lớp 3 á!!!!
Mà bn có vt sai đề bài ko? Mk tính ko ra
\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)
\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)
\(=13\left(1+...+3^7\right)⋮13\)
\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)
\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)
\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)
\(S=4\left(3^2+3^4+3^6+3^8\right)\)
\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)
S = (1 - 3 + 32 - 33) + 34 . (1 - 3 + 32 - 33) + .... + 396 . (1 - 3 + 32 - 33)
S = (-20) + 34 . (-20) +.... + 396 . (-20)
S = (-20) . (1 + 34 +...+ 396)
\(\Rightarrow\)S \(⋮\) 20
(Ko bt có đúng ko)
*KO CHÉP MẠNG*

Ta có: \(S=3^2+3^3+\cdots+3^{2023}\)
\(=\left(3^2+3^3+3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}+3^{12}+3^{13}\right)+\cdots+\left(3^{2018}+3^{2019}+3^{2020}+3^{2021}+3^{2022}+3^{2023}\right)\)
\(=\left(3^2+3^3+\ldots+3^7\right)+3^6\left(3^2+3^3+\cdots+3^7\right)+\cdots+3^{2016}\left(3^2+3^3+\cdots+3^7\right)\)
\(=\left(3^2+3^3+\cdots+3^7\right)\left(1+3^6+\cdots+3^{2016}\right)\)
\(=3276\cdot\left(1+3^6+\cdots+3^{2016}\right)\) ⋮156