Question

URGENTE ME AJUDE DE POSSÍVEL , DESDE JÁ AGRADEÇO.

Se √9ⁿ⁺¹ = 3 e ㏒₂(q - 1) = ¹/₂, então n²+n.q+q² vale:

Answer 1

$\bullet\;\;\sqrt{9^{n+1}}=3\\\\ \sqrt{(3^{2})^{n+1}}=3\\\\ \sqrt{(3^{n+1})^{2}}=3\\\\ 3^{n+1}=3^{1}\\\\ n+1=1\\\\ n=0$

$\bullet\;\;\mathrm{\ell og}_2(q-1)=\dfrac{1}{2}\\\\\ 2^{1/2}=q-1\\\\ \sqrt{2}=q-1\\\\ q=\sqrt{2}+1$

________

Sendo assim,

$n^2+nq+q^2\\\\ =0^2+0\cdot \big(\sqrt{2}+1\big)+\big(\sqrt{2}+1\big)^2\\\\ =\big(\sqrt{2}\big)^2+2\cdot \sqrt{2}\cdot 1+1^2\\\\ =2+2\sqrt{2}+1\\\\ =\boxed{\begin{array}{c}3+2\sqrt{2} \end{array}}$


Bons estudos! :-)