Question

Alguem resolve a questão de numero 12?

Answer 1

Vou representar a expressão de forma abreviada:

$A=\sum_{k=0}^{99}{\dfrac{1}{\sqrt{k+1}+\sqrt{k}}}\\ \\ A=\sum_{k=0}^{99}{\dfrac{\sqrt{k+1}-\sqrt{k}}{ \left(\sqrt{k+1}+\sqrt{k} \right )\cdot \left(\sqrt{k+1}-\sqrt{k} \right )}}\\ \\ A=\sum_{k=0}^{99}{\dfrac{\sqrt{k+1}-\sqrt{k}}{\left(\sqrt{k+1} \right )^{2}-\left(\sqrt{k} \right )^{2}}}\\ \\ A=\sum_{k=0}^{99}{\dfrac{\sqrt{k+1}-\sqrt{k}}{ \left(k+1\right )-k \right )}}\\ \\ A=\sum_{k=0}^{99}{\left(\sqrt{k+1}-\sqrt{k} \right )}\\ \\ \\ A=\left(\sqrt{1}-\sqrt{0} \right )+\left(\sqrt{2}-\sqrt{1} \right )+...+\left(\sqrt{100}-\sqrt{99} \right )\\ \\ A=1+\left(\sqrt{2}-1 \right )+...+\left(\sqrt{100}-\sqrt{99} \right )\\ \\ A=1-1+\sqrt{2}-\sqrt{2}+...+\sqrt{99}-\sqrt{99}+\sqrt{100}$


Todos os termos se anulam, exceto $\sqrt{100}$. Logo,

$A=\sqrt{100}\\ \\ A=10$


Resposta: alternativa $\text{a) }10$.