Question

10.Sa se arate ca numarul S=3+3^2+3^3+...+3^222 este divizibil cu 12

Answer 1

Răspuns:

Explicație pas cu pas:

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$\bf S=3+3^{2}+3^{3}+......+3^{222}$

• Grupam termenii

$\bf S=(3^{1}+3^{2})+3^{2}\cdot(3^{3-2}+3^{4-2})+......+3^{220}\cdot(3^{221-220}+3^{222-220})$

$\bf S=(3+9)+3^{2} \cdot (3^{1}+3^{2})+......+3^{220}\cdot(3^{1}+3^{2})$

$\bf S=12+3^{2} \cdot (3+9)+......+3^{220}\cdot(3+9)$

$\bf S=12+3^{2} \cdot 12+......+3^{220}\cdot 12$

• Dam factor comun 12

$\boxed{\boxed{\bf S=12\cdot (1 +3^{2}+3^{3}+ ......3^{219}+3^{220})~\vdots ~12}}$

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