Ache o 5° termo do Binômio;
a) (4x - 3y) *6 b) (2x + y) *8
Soluções para a tarefa
Respondido por
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Termo geral:
![\boxed{T_{p+1} = \begin{pmatrix}
n\\
p
\end{pmatrix} \cdot a^{p} \cdot x^{n-p}} \boxed{T_{p+1} = \begin{pmatrix}
n\\
p
\end{pmatrix} \cdot a^{p} \cdot x^{n-p}}](https://tex.z-dn.net/?f=%5Cboxed%7BT_%7Bp%2B1%7D+%3D+%5Cbegin%7Bpmatrix%7D%0An%5C%5C+%0Ap%0A%5Cend%7Bpmatrix%7D+%5Ccdot+a%5E%7Bp%7D+%5Ccdot+x%5E%7Bn-p%7D%7D)
a) Para ser o 5° termo, o "T" tem que valer 5. E para isso:
p+1 = 5
p = 5-1
p = 4
Voltando:
![T_{4+1} = \begin{pmatrix}
6\\
4
\end{pmatrix} \cdot (-3y)^{4} \cdot (4x)^{6-4}
\\\\
T_{5} = \frac{6!}{(6-4)! \cdot 4!} \cdot (-3y)^{4} \cdot (4x)^{2}
\\\\
T_{5} = \frac{6!}{2! \cdot 4!} \cdot 81y^{4} \cdot16x^{2}
\\\\
T_{5} = 15 \cdot 81y^{4} \cdot 16x^{2}
\\\\
\boxed{\boxed{T_{5} = 19440y^{4}x^{2}}} T_{4+1} = \begin{pmatrix}
6\\
4
\end{pmatrix} \cdot (-3y)^{4} \cdot (4x)^{6-4}
\\\\
T_{5} = \frac{6!}{(6-4)! \cdot 4!} \cdot (-3y)^{4} \cdot (4x)^{2}
\\\\
T_{5} = \frac{6!}{2! \cdot 4!} \cdot 81y^{4} \cdot16x^{2}
\\\\
T_{5} = 15 \cdot 81y^{4} \cdot 16x^{2}
\\\\
\boxed{\boxed{T_{5} = 19440y^{4}x^{2}}}](https://tex.z-dn.net/?f=T_%7B4%2B1%7D+%3D+%5Cbegin%7Bpmatrix%7D%0A6%5C%5C+%0A4%0A%5Cend%7Bpmatrix%7D+%5Ccdot+%28-3y%29%5E%7B4%7D+%5Ccdot+%284x%29%5E%7B6-4%7D%0A%5C%5C%5C%5C%0AT_%7B5%7D+%3D+%5Cfrac%7B6%21%7D%7B%286-4%29%21+%5Ccdot+4%21%7D+%5Ccdot+%28-3y%29%5E%7B4%7D+%5Ccdot+%284x%29%5E%7B2%7D%0A%5C%5C%5C%5C%0AT_%7B5%7D+%3D+%5Cfrac%7B6%21%7D%7B2%21+%5Ccdot+4%21%7D+%5Ccdot+81y%5E%7B4%7D+%5Ccdot16x%5E%7B2%7D%0A%5C%5C%5C%5C%0AT_%7B5%7D+%3D+15+%5Ccdot+81y%5E%7B4%7D+%5Ccdot+16x%5E%7B2%7D%0A%5C%5C%5C%5C%0A%5Cboxed%7B%5Cboxed%7BT_%7B5%7D+%3D+19440y%5E%7B4%7Dx%5E%7B2%7D%7D%7D)
b)
![T_{4+1} = \begin{pmatrix} 8 \\ 4 \end{pmatrix} \cdot (y)^{4} \cdot (2x)^{8-4}
\\\\
T_{5} = \frac{8!}{(8-4)! \cdot 4!} \cdot (y)^{4} \cdot (2x)^{4}
\\\\
T_{5} = 70 \cdot y^{4} \cdot 16x^{4}
\\\\
\boxed{\boxed{T_{5} = 1120y^{4}x^{4}}} T_{4+1} = \begin{pmatrix} 8 \\ 4 \end{pmatrix} \cdot (y)^{4} \cdot (2x)^{8-4}
\\\\
T_{5} = \frac{8!}{(8-4)! \cdot 4!} \cdot (y)^{4} \cdot (2x)^{4}
\\\\
T_{5} = 70 \cdot y^{4} \cdot 16x^{4}
\\\\
\boxed{\boxed{T_{5} = 1120y^{4}x^{4}}}](https://tex.z-dn.net/?f=T_%7B4%2B1%7D+%3D+%5Cbegin%7Bpmatrix%7D+8+%5C%5C+4+%5Cend%7Bpmatrix%7D+%5Ccdot+%28y%29%5E%7B4%7D+%5Ccdot+%282x%29%5E%7B8-4%7D%0A%5C%5C%5C%5C%0AT_%7B5%7D+%3D+%5Cfrac%7B8%21%7D%7B%288-4%29%21+%5Ccdot+4%21%7D+%5Ccdot+%28y%29%5E%7B4%7D+%5Ccdot+%282x%29%5E%7B4%7D%0A%5C%5C%5C%5C%0AT_%7B5%7D+%3D+70+%5Ccdot+y%5E%7B4%7D+%5Ccdot+16x%5E%7B4%7D%0A%5C%5C%5C%5C%0A%5Cboxed%7B%5Cboxed%7BT_%7B5%7D+%3D+1120y%5E%7B4%7Dx%5E%7B4%7D%7D%7D)
a) Para ser o 5° termo, o "T" tem que valer 5. E para isso:
p+1 = 5
p = 5-1
p = 4
Voltando:
b)
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