Question

como resolver a expressão:
(2√3 + √5)∧5 - (2√3 - √5)∧5

∧ = Elevado!

Answer 1

Lembremos do binômio de Newton:

$\bullet\;\;(2\sqrt{3}+\sqrt{5})^5=\displaystyle\sum_{k=0}^{5} \binom{5}{k}(2\sqrt{3})^{5-k}(\sqrt{5})^k\\\\\\ \bullet\;\;(2\sqrt{3}-\sqrt{5})^5=\sum_{k=0}^{5} \binom{5}{k}(2\sqrt{3})^{5-k}(-\sqrt{5})^k$


Portanto,

$(2\sqrt{3}+\sqrt{5})^5-(2\sqrt{3}-\sqrt{5})^5\\\\=\displaystyle\sum_{k=0}^{5} \binom{5}{k}(2\sqrt{3})^{5-k}(\sqrt{5})^k-\sum_{k=0}^{5} \binom{5}{k}(2\sqrt{3})^{5-k}(-\sqrt{5})^k\\\\\\ =\sum_{k=0}^{5} \left[\binom{5}{k}(2\sqrt{3})^{5-k}(\sqrt{5})^k-\binom{5}{k}(2\sqrt{3})^{5-k}(-\sqrt{5})^k \right]\\\\\\ =\sum_{k=0}^{5} \binom{5}{k}(2\sqrt{3})^{5-k}\left[(\sqrt{5})^k-(-\sqrt{5})^k \right]\\\\\\ =\sum_{k=0}^{5} \binom{5}{k}(2\sqrt{3})^{5-k}(\sqrt{5})^k\left[1^k-(-1)^k \right]~~~~~~\mathbf{(i)}$


Analisemos o termo dentro do somatório acima. Sabemos que

$\bullet\;\;$ Para $k$ par,

$1^k-(-1)^k=1-1=0$

Logo, os termos do somatório se anulam para $k=0,\,2,\,4.$


$\bullet\;\;$ Para $k$ ímpar,

$1^k-(-1)^k=1-(-1)=2$

para $k=1,\,3,\,5.$

_________________

Então, o somatório $\mathbf{(i)}$ se reduz a

$=\displaystyle\binom{5}{1}(2\sqrt{3})^{4}(\sqrt{5})^1\cdot 2+\binom{5}{3}(2\sqrt{3})^{2}(\sqrt{5})^3\cdot 2+\binom{5}{5}(2\sqrt{3})^{0}(\sqrt{5})^5\cdot 2\\\\\\ =5\cdot 144\cdot \sqrt{5}\cdot 2+10\cdot 12\cdot 5\sqrt{5}\cdot 2+1\cdot 1\cdot 25\sqrt{5}\cdot 2\\\\\\ =1\,440\sqrt{5}+1\,200\sqrt{5}+50\sqrt{5}\\\\\\ =\boxed{\begin{array}{c}2\,690\sqrt{5} \end{array}}$


Bons estudos! :-)