Câu hỏi của Nguy Han

Question

Cho A = 3+3 mũ2 + 3 mũ3 + .......+ 3 mũ 2024 .Tìm số dư khi chia A cho 13          b)Chứng tỏ rằng: A = 1+ 4+ 4mũ2 +4mũ3+......+ 4mũ2021 chia hết cho 21.                                               ...

HT.Phong (9A5) · Accepted Answer

Ta có: $A=3+3^2+3^3+...+3^{2024}\
=\left(3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{2022}+3^{2023}+3^{2024}\right)\
=12+3^3\cdot\left(1+3+3^2\right)+3^{2022}\cdot\left(1+3+3^2\right)\
=12+13\cdot\left(3^3+...+3^{2022}\right)$=> A chia 13 dư 12 $A=1+4+4^2+...+4^{2021}\
=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{2019}+4^{2020}+4^{2021}\right)\
=21+4^3\cdot\left(1+4+4^2\right)+...+4^{2019}\cdot\left(1+4+4^2\right)\
=21+4^3\cdot21+...+4^{2019}\cdot21\
=21\cdot\left(1+4^3+...+4^{2019}\right)$Câu trả lời: Ta có: $A=3+3^2+3^3+...+3^{2024}\
=\left(3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{2022}+3^{2023}+3^{2024}\right)\
=12+3^3\cdot\left(1+3+3^2\right)+3^{2022}\cdot\left(1+3+3^2\right)\
=12+13\cdot\left(3^3+...+3^{2022}\right)$=> A chia 13 dư 12 $A=1+4+4^2+...+4^{2021}\
=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{2019}+4^{2020}+4^{2021}\right)\
=21+4^3\cdot\left(1+4+4^2\right)+...+4^{2019}\cdot\left(1+4+4^2\right)\
=21+4^3\cdot21+...+4^{2019}\cdot21\
=21\cdot\left(1+4^3+...+4^{2019}\right)$

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