Toán Tìm n ∈ N để $(2n+1)^3 +1$ chia hết cho $2^{2021}$ 14/07/2021 By Cora Tìm n ∈ N để $(2n+1)^3 +1$ chia hết cho $2^{2021}$
Đề thi vào 10 chuyên PTNK năm 2021-2022 $\begin{array}{l} {\left( {2n + 1} \right)^3} + 1 \vdots {2^{2021}}\\ \Leftrightarrow \left( {2n + 1 + 1} \right)\left[ {{{\left( {2n + 1} \right)}^2} – \left( {2n + 1} \right) + 1} \right] \vdots {2^{2021}}\\ \Leftrightarrow 2\left( {n + 1} \right)\left[ {4{n^2} + 2n + 1} \right] \vdots {2^{2021}}\\ \Leftrightarrow \left( {n + 1} \right)\left( {4{n^2} + 2n + 1} \right) \vdots {2^{2020}}\\ \Rightarrow n + 1 \vdots {2^{2020}}\left( {4{n^2} + 2n + 1 \equiv 1\left( {\bmod 2} \right)} \right)\\ \Rightarrow n + 1 = {2^{2020}}k\\ \Rightarrow n = {2^{2020}}k – 1\left( {k \in \mathbb{Z}} \right) \end{array}$ Trả lời
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Đề thi vào 10 chuyên PTNK năm 2021-2022
$\begin{array}{l} {\left( {2n + 1} \right)^3} + 1 \vdots {2^{2021}}\\ \Leftrightarrow \left( {2n + 1 + 1} \right)\left[ {{{\left( {2n + 1} \right)}^2} – \left( {2n + 1} \right) + 1} \right] \vdots {2^{2021}}\\ \Leftrightarrow 2\left( {n + 1} \right)\left[ {4{n^2} + 2n + 1} \right] \vdots {2^{2021}}\\ \Leftrightarrow \left( {n + 1} \right)\left( {4{n^2} + 2n + 1} \right) \vdots {2^{2020}}\\ \Rightarrow n + 1 \vdots {2^{2020}}\left( {4{n^2} + 2n + 1 \equiv 1\left( {\bmod 2} \right)} \right)\\ \Rightarrow n + 1 = {2^{2020}}k\\ \Rightarrow n = {2^{2020}}k – 1\left( {k \in \mathbb{Z}} \right) \end{array}$