Question

O valor de... é?
Me mostre a resolução, por favor!!!

Answer 1

Usando o Teorema da soma de números binomiais 

$\sum_{p=1}^{8}\binom{8}{p}~\to~\binom{8}{1}+\binom{8}{2}...+..+\binom{8}{8}$

Pelo teorema

$\boxed{\binom{n}{0}+\binom{n}{1}+...+\binom{n}{n}=2^n}$

Aplicando a definição e subtraindo o primeiro termo


$\underbrace{\binom{8}{0} +\binom{8}{1}+\binom{8}{2}...+..+\binom{8}{8}}-\underbrace{\binom{8}{0}}\\ \\2^8-1\\ \\256-1\\ \\\boxed{\boxed{\therefore255}}$

Answer 2

$\displaystyle\sum_{p=1}^{8}~\dbinom{8}{p}=\dbinom{8}{1}+\dbinom{8}{2}+\dots+\dbinom{8}{8}$

Veja que:

$\dbinom{8}{1}=\dfrac{8!}{1!\cdot7!}=8$

$\dbinom{8}{2}=\dfrac{8!}{2!\cdot6!}=28$

$\dbinom{8}{3}=\dfrac{8!}{3!\cdot5!}=56$

$\dbinom{8}{4}=\dfrac{8!}{4!\cdot4!}=70$

Mas, $\dbinom{n}{p}=\dbinom{n}{n-p}$.

Com isso,

$\dbinom{8}{1}=\dbinom{8}{7}$,

$\dbinom{8}{2}=\dbinom{8}{6}$

$\dbinom{8}{3}=\dbinom{8}{5}$

Assim, $\dbinom{8}{5}=56$, $\dbinom{8}{6}=28$, $\dbinom{8}{7}=8$.

Além disso, $\dbinom{8}{8}=\dfrac{8!}{8!\cdot0!}=1$.

Logo:

$\displaystyle\sum_{p=1}^{8}~\dbinom{8}{p}=\dbinom{8}{1}+\dbinom{8}{2}+\dots+\dbinom{8}{8}=8+28+56+70+56+28+8+1=255$

Letra B