Toán Lớp 8: CM rằng n lẻ thì:
a, n^2 + 4n + 3 chia hết cho 8
b, n^3 + 3n^2 – n- 3 chia hết cho 48
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Toán Lớp 8: CM rằng n lẻ thì:
a, n^2 + 4n + 3 chia hết cho 8
b, n^3 + 3n^2 – n- 3 chia hết cho 48
Comments ( 1 )
a){n^2} + 4n + 3\\
= {\left( {2k + 1} \right)^2} + 4.\left( {2k + 1} \right) + 3\\
= 4{k^2} + 4k + 1 + 8k + 4 + 3\\
= 4{k^2} + 12k + 8\\
= 4.\left( {{k^2} + 3k + 2} \right)\\
= 4.\left( {{k^2} + 2k + k + 2} \right)\\
= 4.\left( {k + 1} \right).\left( {k + 2} \right)
\end{array}$
\Leftrightarrow \left( {k + 1} \right)\left( {k + 2} \right) \vdots 2\\
\Leftrightarrow 4.\left( {k + 1} \right)\left( {k + 2} \right) \vdots 8\\
\Leftrightarrow {n^2} + 4n + 3 \vdots 8\\
b){n^3} + 3{n^2} – n – 3\\
= {\left( {2k + 1} \right)^3} + 3.{\left( {2k + 1} \right)^2} – \left( {2k + 1} \right) – 3\\
= 8{k^3} + 12{k^2} + 6k + 1 + 3.\left( {4{k^2} + 4k + 1} \right)\\
– 2k – 1 – 3\\
= 8{k^3} + 12{k^2} + 4k – 3 + 12{k^2} + 12k + 3\\
= 8{k^3} + 24{k^2} + 16k\\
= 8.k\left( {{k^2} + 3k + 2} \right)\\
= 8k.\left( {k + 1} \right).\left( {k + 2} \right)
\end{array}$
\Leftrightarrow k\left( {k + 1} \right)\left( {k + 2} \right) \vdots 6\\
\Leftrightarrow 8k\left( {k + 1} \right)\left( {k + 2} \right) \vdots 48\\
\Leftrightarrow {n^3} + 3{n^2} – n – 3 \vdots 48
\end{array}$