Question

Alguém poderia me ajudar? com o desenvolvimento do binômio (2a-3b)^7 pela formula:
(a+b)^n=Σ(n p) a^n-pb^p! 

Answer 1

$(x+y)^{n}=\sum\limits_{p=0}^{n}~\left(\begin{array}{c}n\\p\end{array}\right)*x^{n-p}*y^{p}$

Lembrando que:

$\left(\begin{array}{c}n\\p\end{array}\right)=C_{n,p}=\dfrac{n!}{p!(n-p)!}$
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$(2a-3b)^{7}=\sum\limits_{p=0}^{7}~\left(\begin{array}{c}7\\p\end{array}\right)*(2a)^{7-p}*(-3b)^{p}$

Vou dividir esse somatório em 8 partes, já que não daria pra deixar tudo junto no LaTeX e também ficaria muito desorganizado

$(2a-3b)^{7}=u_{0}+u_{1}+u_{2}+u_{3}+u_{4}+u_{5}+u_{6}+u_{7}$

Os índices indicam os valores de p

$u_{0}=\left(\begin{array}{c}7\\0\end{array}\right)*(2a)^{7-0}*(-3b)^{0}=1*(2a)^{7}=2^{7}a^{7}=128a^{7}\\\\u_{1}=\left(\begin{array}{c}7\\1\end{array}\right)*(2a)^{7-1}*(-3b)^{1}=7*(2a)^{6}*(-3b)=-1344a^{6}b$

$u_{2}=\left(\begin{array}{c}7\\2\end{array}\right)*(2a)^{7-2}*(-3b)^{2}=21*(2a)^{5}*9b^{2}=6048a^{5}b^{2}\\\\u_{3}=\left(\begin{array}{c}7\\3\end{array}\right)*(2a)^{7-3}*(-3b)^{3}=35*(2a)^{4}*(-27b^{3})=-15120a^{4}b^{3}$

$u_{4}=\left(\begin{array}{c}7\\4\end{array}\right)*(2a)^{7-4}*(-3b)^{4}=35*(2a)^{3}*(81b^{4})=22680a^{3}b^{4}\\\\u_{5}=\left(\begin{array}{c}7\\5\end{array}\right)*(2a)^{7-5}*(-3b)^{5}=21*(2a)^{2}*(-243b^{5})=-20412a^{2}b^{5}$

$u_{6}=\left(\begin{array}{c}7\\6\end{array}\right)*(2a)^{7-6}*(-3b)^{6}=7*(2a)^{1}*(729b^{6})=10206ab^{6}\\\\u_{7}=\left(\begin{array}{c}7\\7\end{array}\right)*(2a)^{7-7}*(-3b)^{7}=1*(2a)^{0}*(-2187b^{7})=-2187b^{7}$
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Finalmente:

$(2a-3b)^{7}=u_{0}+u_{1}+u_{2}+u_{3}+u_{4}+u_{5}+u_{6}+u_{7}$

(2a - 3b)⁷ = 128a⁷ - 1344a⁶b + 6048a⁵b² - 15120a⁴b³ + 22680a³b⁴ - 20412a²b⁵ + 10206ab⁶ - 2187b⁷