Question

Arătați că numărul A = 8^n+1 × 2^n+3 + 3 × 8^n × 2^n+1 + 15 × 8^n× 2^n+1 este pătrat perfect
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Answer 1

Răspuns:

Sper ca te am ajutat!!!!!!

Answer 2

Răspuns: $\color{Crimson}\large \boxed{\bf A = (2^{2n}\cdot 10)^{2}\longrightarrow patrat~perfect}$

Explicație pas cu pas:

$\large \bf A = 8^{n+1}\cdot 2^{n+3} + 3 \cdot 8^{n} \cdot 2^{n+1}+ 15 \cdot 8^{n}\cdot 2^{n+1}$

$\large \bf A = (2^{3} )^{n+1}\cdot 2^{n+3} + 3 \cdot (2^{3} )^{n} \cdot 2^{n+1}+ 15 \cdot (2^{3} )^{n}\cdot 2^{n+1}$

$\large \bf A = 2^{3n+3}\cdot 2^{n+3} + 3 \cdot 2^{3n} \cdot 2^{n+1}+ 15 \cdot 2^{3n}\cdot 2^{n+1}$

$\large \bf A = 2^{3n+3+n+3}+ 3 \cdot 2^{3n+n+1} + 15 \cdot 2^{3n+n+1}$

$\large \bf A = 2^{4n+6}+ 3 \cdot 2^{4n+1} + 15 \cdot 2^{4n+1}$

$\large \bf A = 2^{4n}\cdot (2^{6}+ 3 \cdot 2^{1} + 15 \cdot 2^{1})$

$\large \bf A = 2^{4n}\cdot (64+ 6 + 15 \cdot 2)$

$\large \bf A = 2^{4n}\cdot (64+ 6 + 30)$

$\large \bf A = 2^{4n}\cdot 100$

$\large \bf A = 2^{4n}\cdot 10^{2}$

$\color{red}\large \boxed{\bf A = (2^{2n}\cdot 10)^{2}\longrightarrow patrat~perfect}$

==pav38==