Question

Qual é o coeficiente de x^6 na expansão de (2x+1)^8?

a) 1792
b) 16
c) 256
d) 1120
e) 1024

Answer 1

Expansão / Binômio de Newton :

$\displaystyle \sf (A+B)^{n} = \sum_{k=0 }^{n}{n\choose p}\cdot A^{(n-p)}. B^{p}$

Termo geral do Binômio de Newton  :

$\displaystyle \sf T_{p+1} = {n\choose p }\cdot A^{(n-p)}\cdot B^{p}$

Queremos o coeficiente de $\sf x^{6}$ na expansão $\sf (2x+1)^{8}$. Então :

$\displaystyle \sf T_{p+1} = {8\choose p }\cdot (2x)^{8-p}\cdot 1^{p } \\\\\\ T_{p+1}={8\choose p}\cdot x^{(8-p)}\cdot 2^{(8-p)} \\\\\\ \underline{devemos\ ter} : \\\\ x^{(8-p)} = x^6 \to 8-p=6 \to \boxed{\sf p = 2} \\\\\ \underline{Da{\'i}}}: \\\\ T_{2+1} = {8\choose 2}\cdot x^{(8-2)}\cdot 2^{(8-2)} \\\\\\ T_{3}=\frac{8!}{(8-2)!\cdot 2! }\cdot x^{6}\cdot 2^{6} \\\\\\ T_3=\frac{8\cdot7\cdot 6! }{6!\cdot 2!}\cdot x^6\cdot 64 \\\\\\ T_3 = 28\cdot 64 \cdot x^6 \\\\ \boxed{\sf T_{3} = 1792\cdot x^6 }$

Coeficiente = 1792