Question

Determinar o coeficiente de x²⁰ no desenvolvimento de (x⁴ + 3)⁶​

Answer 1

Resposta:

o coeficiente é 18

Explicação passo a passo:

$(x^{4} + 3)^{6}$ = ($x^{4}$ + 3)². ($x^{4}$ + 3)². ($x^{4}$ + 3)² =

(($x^{4}$)² + 2.$x^{4}$.3 + 3²).(($x^{4}$)² + 2.$x^{4}$.3 + 3²).(($x^{4}$)² + 2.$x^{4}$.3 + 3²) =

($x^{8}$ + 6$x^{4}$ + 9).($x^{8}$ + 6$x^{4}$ + 9).($x^{8}$ + 6$x^{4}$ + 9) =

($x^{16}$ + 6$x^{12}$ + 9$x^{8}$ + 6$x^{12}$ + 36$x^{8}$ + 64$x^{4}$ + 9$x^{8}$ + 54$x^{4}$ + 81).($x^{8}$ + 6$x^{4}$ + 9) =

($x^{16}$ + 12$x^{12}$ + 54$x^{8}$ + 118$x^{4}$ + 81).($x^{8}$ + 6$x^{4}$ + 9) =

$x^{24}$ + 6$x^{20}$ + 9$x^{16}$ + 12$x^{20}$ + 72$x^{16}$ + 108$x^{12}$ + 54$x^{16}$ + 324$x^{12}$ + 486$x^{8}$ + 118$x^{12}$ + 708$x^{8}$ + 1062$x^{4}$ + 81$x^{8}$ + 486$x^{4}$ + 729 =

$x^{24}$ + 18$x^{20}$ + 135$x^{16}$ + 550$x^{12}$ + 1275$x^{8}$ + 1548$x^{4}$ + 729

Answer 2

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\sf (x^4 + 3)^6$

$\sf T_{p+1} = \binom{n}{p}\:.\:x^{n - p}\:.\:y^p$

$\sf T_{p+1} = \binom{6}{p}\:.\:(x^4)^{6 - p}\:.\:3^p$

$\sf T_{p+1} = \binom{6}{p}\:.\:x^{24 - 4p}\:.\:3^p$

$\textsf{24 - 4p = 20}$

$\textsf{4p = 4}$

$\textsf{p = 1}$

$\sf \binom{6}{1}\:.\:3^1 = \left(\dfrac{6!}{1!\:.\:(6-1)!}\right).3 = 6\:.\:3$

$\boxed{\boxed{\sf 18}}\leftarrow\textsf{coeficiente de }\sf x^{20}$