Question

Alguém pode me explicar passo a passo essa aqui

Answer 1

Binômio de Newton:

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

Termo geral do Binômio de Newton:

$\boxed{\boxed{T_{k+1}=C_{n,k}\cdot a^{n-k}\cdot b^{k}}}$

OBS:

$\boxed{\boxed{C_{n,k}=\dfrac{n!}{k!(n-k)!}}}}$
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$\left(x^{3}-\dfrac{1}{x^{2}}\right)^{10}=(x^{3}-x^{-2})^{10}$

Logo, temos o Binômio:

$(x^{3}-x^{-2})^{10}=\displaystyle\sum\limits_{k=0}^{10}C_{10,k}\cdot(x^{3})^{10-k}\cdot(x^{-2})^{k}\\\\\\(x^{3}-x^{-2})^{10}=\sum\limits_{k=0}^{10}C_{10,k}\cdot x^{30-3k}\cdot x^{-2k}\\\\\\(x^{3}-x^{-2})^{10}=\sum\limits_{k=0}^{10}C_{10,k}\cdot x^{30-3k-2k}\\\\\\\boxed{\boxed{(x^{3}-x^{-2})^{10}=\sum\limits_{k=0}^{10}C_{10,k}\cdot x^{30-5k}}}$

O termo independente será o termo que possui 0 no expoente de x.

Portanto:

$30-5k=0~~~\therefore~~\boxed{\boxed{k=6}}$

Achando o termo com k = 6 (sétimo termo):

$T_{6+1}=C_{10,6}\cdot x^{30-5\cdot6}\\\\\\T_{7}=\dfrac{10!}{6!(10-6)!}\cdot x^{0}\\\\\\T_{7}=\dfrac{10\cdot9\cdot8\cdot7\cdot6!}{6!\cdot4!}\\\\\\T_{7}=\dfrac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2\cdot1}\\\\\\T_{7}=10\cdot3\cdot7\\\\\\\boxed{\boxed{T_{7}=210}}$

Letra B