Toán tìm số tự nhiên n để sao cho (2n+1) ³+1 chia hết cho $2^{2021}$ 22/07/2021 By Madeline tìm số tự nhiên n để sao cho (2n+1) ³+1 chia hết cho $2^{2021}$
$\begin{array}{l} {\left( {2n + 1} \right)^3} + 1 \vdots {2^{2021}}\\ \Leftrightarrow \left( {2n + 1 + 1} \right)\left[ {{{\left( {2n + 1} \right)}^2} – \left( {2n + 1} \right) + 1} \right] \vdots {2^{2021}}\\ \Leftrightarrow \left( {2n + 2} \right)\left( {4{n^2} + 2n + 1} \right) \vdots {2^{2021}}\\ \Leftrightarrow \left( {n + 1} \right)\left( {4{n^2} + 2n + 1} \right) \vdots {2^{2020}}\\ \Leftrightarrow \left( {n + 1} \right)\left( {4{n^2} + 2n + 1} \right) \vdots {2^{2020}}\\ \Leftrightarrow n + 1 \vdots {2^{2020}}\left( {4{n^2} + 2n + 1 \equiv 1\left( {\bmod 2} \right)} \right)\\ \Leftrightarrow n + 1 = {2^{2020}}k\\ \Leftrightarrow n = {2^{2020}}k – 1\left( {k \in \mathbb{Z} + } \right) \end{array}$ Trả lời
`(2n+1)^3+1` `=(2n+1+1)((2n+1)^2-(2n+1)+1)` `=2(n+1)(4n^2+4n+1-2n+1-1)` `=2(n+1)(4n^2+2n+1)\vdots2^(2021)` `=(n+1)(4n^2+2n+1)\vdots2^(2020)` `⇔n+1\vdots2^(2020)`(vì`4n^2+2n+1\cancel{vdots}2^(2021)`) `⇔n` lẻ `n>2^(2020)` `⇒n` có dạng`2^(2020)x-1` Trả lời
$\begin{array}{l} {\left( {2n + 1} \right)^3} + 1 \vdots {2^{2021}}\\ \Leftrightarrow \left( {2n + 1 + 1} \right)\left[ {{{\left( {2n + 1} \right)}^2} – \left( {2n + 1} \right) + 1} \right] \vdots {2^{2021}}\\ \Leftrightarrow \left( {2n + 2} \right)\left( {4{n^2} + 2n + 1} \right) \vdots {2^{2021}}\\ \Leftrightarrow \left( {n + 1} \right)\left( {4{n^2} + 2n + 1} \right) \vdots {2^{2020}}\\ \Leftrightarrow \left( {n + 1} \right)\left( {4{n^2} + 2n + 1} \right) \vdots {2^{2020}}\\ \Leftrightarrow n + 1 \vdots {2^{2020}}\left( {4{n^2} + 2n + 1 \equiv 1\left( {\bmod 2} \right)} \right)\\ \Leftrightarrow n + 1 = {2^{2020}}k\\ \Leftrightarrow n = {2^{2020}}k – 1\left( {k \in \mathbb{Z} + } \right) \end{array}$
`(2n+1)^3+1`
`=(2n+1+1)((2n+1)^2-(2n+1)+1)`
`=2(n+1)(4n^2+4n+1-2n+1-1)`
`=2(n+1)(4n^2+2n+1)\vdots2^(2021)`
`=(n+1)(4n^2+2n+1)\vdots2^(2020)`
`⇔n+1\vdots2^(2020)`(vì`4n^2+2n+1\cancel{vdots}2^(2021)`)
`⇔n` lẻ
`n>2^(2020)`
`⇒n` có dạng`2^(2020)x-1`