Question

Calcule o valor da expressão (1 + √a)⁴ − (1 − √a)⁴ , usando o teorema binomial.

Answer 1

$\boxed{(a+b)^n=\sum\limits_{p=0}^{n}\binom{n}{p}a^{n-p}b^p}$

Chamaremos a expressão de $K$:

$K=(1+\sqrt{a})^4-(1-\sqrt{a})^4$

$K=\sum\limits_{i=0}^{4}{\dbinom{4}{i}1^{4-i}(\sqrt{a})^i}-\sum\limits_{j=0}^{4}{\dbinom{4}{j}1^{4-j}(-\sqrt{a})^{j}}$

$K=\sum\limits_{i=0}^{4}{\dbinom{4}{i}a^{\tfrac{i}{2}}}-\sum\limits_{j=0}^{4}{\dbinom{4}{j}(-1)^ja^{\tfrac{j}{2}}}$

$K=\bigg[\dbinom{4}{0}a^0+\dbinom{4}{1}a^{\frac{1}{2}}+\dbinom{4}{2}a^{\frac{2}{2}}+\dbinom{4}{3}a^{\frac{3}{2}}+\dbinom{4}{4}a^{\frac{4}{2}}\bigg]-\\\\ \bigg[\dbinom{4}{0}(-1)^0a^0+\dbinom{4}{1}(-1)^1a^{\frac{1}{2}}+\dbinom{4}{2}(-1)^2a^{\frac{2}{2}}+\dbinom{4}{3}(-1)^3a^{\frac{3}{2}}+\dbinom{4}{4}(-1)^4a^{\frac{4}{2}}\bigg]$

$K=\bigg[1+4a^{\frac{1}{2}}+6a+4a^{\frac{3}{2}}+a^2\bigg]-\bigg[1-4a^{\frac{1}{2}}+6a-4a^{\frac{3}{2}}+a^2\bigg]$

$K=1+4a^{\frac{1}{2}}+6a+4a^{\frac{3}{2}}+a^2-1+4a^{\frac{1}{2}}-6a+4a^{\frac{3}{2}}-a^2$

$K=8a^{\frac{1}{2}}+8a^{\frac{3}{2}}$

$K=8\sqrt{a}+8a\sqrt{a}$

$\boxed{K=8\sqrt{a}(a+1)}$