Question

Demonstrați ca numarul 4 la putea (2n) +2 la puterea (2n+1) +1, n€N, este patrat perfect

Answer 1

${4}^{2n} + {2}^{2n + 1} + 1$

$= {( {2}^{2}) }^{2n} + {2}^{2n} \times 2 + 1$

$= { ({2}^{2n} )}^{2} + {2}^{2n} \times 2 + 1$

$= {( {2}^{2n} + 1)}^{2} = > p.p$

Formulă :

${(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$

$a = {2}^{2n}$

$b = 1$