Question

Calcule o termo independente de x (aquele em que o expoente de x é igual a zero) no desenvolvimento (x+1/x) elevado a 8.

Answer 1

Binômio de Newton:

$\boxed{\boxed{\big(a+b\big)^{n}=\sum\limits_{k=0}^{n}\binom{n}{k}\,a^{k}\cdot b^{n-k}}}$

Onde $\binom{n}{k}=\frac{n!}{k!(n-k)!}$
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Para $a=x$, $b=\frac{1}{x}$ e $n=8$, temos

$\bigg(x+\dfrac{1}{x}\bigg)^{8}=\displaystyle\sum_{k=0}^{8}\binom{8}{k}x^{k}\bigg(\frac{1}{x}\bigg)^{8-k}\\\\\\\bigg(x+\dfrac{1}{x}\bigg)^{8}=\sum_{k=0}^{8}\binom{8}{k}x^{k}\cdot\dfrac{1}{x^{8-k}}\\\\\\\bigg(x+\dfrac{1}{x}\bigg)^{8}=\sum_{k=0}^{8}\binom{8}{k}x^{k-(8-k)}\\\\\\\bigg(x+\dfrac{1}{x}\bigg)^{8}=\sum_{k=0}^{8}\binom{8}{k}x^{2k-8}$

Encontrando $k$ tal que $x^{2k-8}=x^{0}$:

$x^{2k-8}=x^{0}\\\\2k-8=0\\\\2k=8\\\\\boxed{\boxed{k=4}}$

Portanto, o termo independente de $x$ será o quinto termo do desenvolvimento do binômio:

$T_{5}=T_{4+1}=\displaystyle\binom{8}{4}\cdot x^{0}\\\\\\T_{5}=\dfrac{8!}{4!(8-4)!}\\\\\\T_{5}=\dfrac{8\cdot7\cdot6\cdot5\cdot4!}{4!\cdot4!}\\\\\\T_{5}=\dfrac{8\cdot7\cdot6\cdot5}{4\cdot3\cdot2}\\\\\\T_{5}=7\cdot2\cdot5\\\\\\\boxed{\boxed{T_{5}=70}}$