CMR: •25^(n+2)-2.25^n+625^(1/2(n+1)) luôn luôn chia hết cho 648 với mọi n 18/07/2021 Bởi Rose CMR: •25^(n+2)-2.25^n+625^(1/2(n+1)) luôn luôn chia hết cho 648 với mọi n
`25^{n+2)-2.25^n+625^{1/2(n+1)}` `⇔25^{n+2}-2.25^n+(25^2)^{1/2(n+1)}` `⇔25^{n+2}-2.25^n+25^{n+1}` `⇔25^n (25^2-2+25^1)` `⇔25^ .648` `⇒ 25^{n+2)-2.25^n+625^{1/2(n+1)}\vdots648` Bình luận
Giải thích các bước giải: Ta có: \(\begin{array}{l}{25^{n + 2}} – {2.25^n} + {625^{\frac{1}{2}\left( {n + 1} \right)}}\\ = {25^2}{.25^n} – {2.25^n} + {\left( {{{25}^2}} \right)^{\frac{1}{2}\left( {n + 1} \right)}}\\ = {25^2}{.25^n} – {2.25^n} + {25^{n + 1}}\\ = {25^2}{.25^n} – {2.25^n} + {25.25^n}\\ = {25^n}\left( {{{25}^2} – 2 + 25} \right)\\ = {25^n}.648 \vdots 648\end{array}\) Bình luận
`25^{n+2)-2.25^n+625^{1/2(n+1)}`
`⇔25^{n+2}-2.25^n+(25^2)^{1/2(n+1)}`
`⇔25^{n+2}-2.25^n+25^{n+1}`
`⇔25^n (25^2-2+25^1)`
`⇔25^ .648`
`⇒ 25^{n+2)-2.25^n+625^{1/2(n+1)}\vdots648`
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{25^{n + 2}} – {2.25^n} + {625^{\frac{1}{2}\left( {n + 1} \right)}}\\
= {25^2}{.25^n} – {2.25^n} + {\left( {{{25}^2}} \right)^{\frac{1}{2}\left( {n + 1} \right)}}\\
= {25^2}{.25^n} – {2.25^n} + {25^{n + 1}}\\
= {25^2}{.25^n} – {2.25^n} + {25.25^n}\\
= {25^n}\left( {{{25}^2} – 2 + 25} \right)\\
= {25^n}.648 \vdots 648
\end{array}\)