Binomio de newton ( termo geral ) :
Ache o valor de a de modo que x^5 seja igual ao de x^15 no desenvolvimento de [ 2x^2 +( a/x3) ]^10
Lukyo:
Por acaso seria o coeficiente de x^5 que deve ser igual ao de x^15?
Respostas
respondido por:
4
Caso esteja pelo app, experimente abrir pelo navegador: https://brainly.com.br/tarefa/8605823
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Encontrar o valor de a, de modo que os coeficientes do termo em
seja igual ao coeficiente de
no desenvolvimento de
![\mathsf{\left(2x^2+\dfrac{a}{x^3}\right)^{10}} \mathsf{\left(2x^2+\dfrac{a}{x^3}\right)^{10}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cleft%282x%5E2%2B%5Cdfrac%7Ba%7D%7Bx%5E3%7D%5Cright%29%5E%7B10%7D%7D)
——————————
• Encontrar a fórmula do termo geral do binômio.
No desenvolvimento de
![\mathsf{(v+w)^n} \mathsf{(v+w)^n}](https://tex.z-dn.net/?f=%5Cmathsf%7B%28v%2Bw%29%5En%7D)
o termo da posição p + 1 é dado por
![\mathsf{t_{p+1}=\dbinom{n}{p}\cdot v^{n-p}\cdot w^p}\\\\\\ \mathsf{t_{p+1}=\dfrac{n!}{p!\cdot (n-p)!}\cdot v^{n-p}\cdot w^p} \mathsf{t_{p+1}=\dbinom{n}{p}\cdot v^{n-p}\cdot w^p}\\\\\\ \mathsf{t_{p+1}=\dfrac{n!}{p!\cdot (n-p)!}\cdot v^{n-p}\cdot w^p}](https://tex.z-dn.net/?f=%5Cmathsf%7Bt_%7Bp%2B1%7D%3D%5Cdbinom%7Bn%7D%7Bp%7D%5Ccdot+v%5E%7Bn-p%7D%5Ccdot+w%5Ep%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bt_%7Bp%2B1%7D%3D%5Cdfrac%7Bn%21%7D%7Bp%21%5Ccdot+%28n-p%29%21%7D%5Ccdot+v%5E%7Bn-p%7D%5Ccdot+w%5Ep%7D)
com 0 ≤ p ≤ n.
Para o binômio dado,
![\mathsf{\left(2x^2+\dfrac{a}{x^3}\right)^{10}}\\\\\\ =\mathsf{\left(2x^2+ax^{-3}\right)^{10}} \mathsf{\left(2x^2+\dfrac{a}{x^3}\right)^{10}}\\\\\\ =\mathsf{\left(2x^2+ax^{-3}\right)^{10}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cleft%282x%5E2%2B%5Cdfrac%7Ba%7D%7Bx%5E3%7D%5Cright%29%5E%7B10%7D%7D%5C%5C%5C%5C%5C%5C+%3D%5Cmathsf%7B%5Cleft%282x%5E2%2Bax%5E%7B-3%7D%5Cright%29%5E%7B10%7D%7D)
temos![\mathsf{v=2x^2,~~w=ax^{-3},~~n=10.} \mathsf{v=2x^2,~~w=ax^{-3},~~n=10.}](https://tex.z-dn.net/?f=%5Cmathsf%7Bv%3D2x%5E2%2C%7E%7Ew%3Dax%5E%7B-3%7D%2C%7E%7En%3D10.%7D)
A fórmula do termo geral para este binômio é
![\mathsf{t_{p+1}=\dbinom{10}{p}\cdot (2x^2)^{10-p}\cdot (ax^{-3})^p}\\\\\\ \mathsf{t_{p+1}=\dbinom{10}{p}\cdot 2^{10-p}\cdot (x^2)^{10-p}\cdot a^p\cdot (x^{-3})^p}\\\\\\ \mathsf{t_{p+1}=\dbinom{10}{p}\cdot 2^{10-p}\cdot x^{2(10-p)}\cdot a^p\cdot x^{-3p}}\\\\\\ \mathsf{t_{p+1}=\dbinom{10}{p}\cdot 2^{10-p}\cdot a^p\cdot x^{2(10-p)-3p}} \mathsf{t_{p+1}=\dbinom{10}{p}\cdot (2x^2)^{10-p}\cdot (ax^{-3})^p}\\\\\\ \mathsf{t_{p+1}=\dbinom{10}{p}\cdot 2^{10-p}\cdot (x^2)^{10-p}\cdot a^p\cdot (x^{-3})^p}\\\\\\ \mathsf{t_{p+1}=\dbinom{10}{p}\cdot 2^{10-p}\cdot x^{2(10-p)}\cdot a^p\cdot x^{-3p}}\\\\\\ \mathsf{t_{p+1}=\dbinom{10}{p}\cdot 2^{10-p}\cdot a^p\cdot x^{2(10-p)-3p}}](https://tex.z-dn.net/?f=%5Cmathsf%7Bt_%7Bp%2B1%7D%3D%5Cdbinom%7B10%7D%7Bp%7D%5Ccdot+%282x%5E2%29%5E%7B10-p%7D%5Ccdot+%28ax%5E%7B-3%7D%29%5Ep%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bt_%7Bp%2B1%7D%3D%5Cdbinom%7B10%7D%7Bp%7D%5Ccdot+2%5E%7B10-p%7D%5Ccdot+%28x%5E2%29%5E%7B10-p%7D%5Ccdot+a%5Ep%5Ccdot+%28x%5E%7B-3%7D%29%5Ep%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bt_%7Bp%2B1%7D%3D%5Cdbinom%7B10%7D%7Bp%7D%5Ccdot+2%5E%7B10-p%7D%5Ccdot+x%5E%7B2%2810-p%29%7D%5Ccdot+a%5Ep%5Ccdot+x%5E%7B-3p%7D%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bt_%7Bp%2B1%7D%3D%5Cdbinom%7B10%7D%7Bp%7D%5Ccdot+2%5E%7B10-p%7D%5Ccdot+a%5Ep%5Ccdot+x%5E%7B2%2810-p%29-3p%7D%7D)
![\mathsf{t_{p+1}=\dbinom{10}{p}\cdot 2^{10-p}\cdot a^p\cdot x^{2\,\cdot\, 10-2p-3p}}\\\\\\ \mathsf{t_{p+1}=\dbinom{10}{p}\cdot 2^{10-p}\cdot a^p\cdot x^{20-5p}}\qquad\quad\checkmark \mathsf{t_{p+1}=\dbinom{10}{p}\cdot 2^{10-p}\cdot a^p\cdot x^{2\,\cdot\, 10-2p-3p}}\\\\\\ \mathsf{t_{p+1}=\dbinom{10}{p}\cdot 2^{10-p}\cdot a^p\cdot x^{20-5p}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%5Cmathsf%7Bt_%7Bp%2B1%7D%3D%5Cdbinom%7B10%7D%7Bp%7D%5Ccdot+2%5E%7B10-p%7D%5Ccdot+a%5Ep%5Ccdot+x%5E%7B2%5C%2C%5Ccdot%5C%2C+10-2p-3p%7D%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bt_%7Bp%2B1%7D%3D%5Cdbinom%7B10%7D%7Bp%7D%5Ccdot+2%5E%7B10-p%7D%5Ccdot+a%5Ep%5Ccdot+x%5E%7B20-5p%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
Perceba que o expoente de x na posição p + 1 é sempre dado por 20 – 5p.
Sendo assim, vamos encontrar em que posições ocorrem os termos em
e ![\mathsf{x^{15}.} \mathsf{x^{15}.}](https://tex.z-dn.net/?f=%5Cmathsf%7Bx%5E%7B15%7D.%7D)
• Termo em![\mathsf{x^5}: \mathsf{x^5}:](https://tex.z-dn.net/?f=%5Cmathsf%7Bx%5E5%7D%3A)
![\mathsf{x^{20-5p}=x^5}\\\\ \mathsf{20-5p=5}\\\\ \mathsf{5p=20-5}\\\\ \mathsf{5p=15}\\\\ \mathsf{p=\dfrac{15}{5}} \mathsf{x^{20-5p}=x^5}\\\\ \mathsf{20-5p=5}\\\\ \mathsf{5p=20-5}\\\\ \mathsf{5p=15}\\\\ \mathsf{p=\dfrac{15}{5}}](https://tex.z-dn.net/?f=%5Cmathsf%7Bx%5E%7B20-5p%7D%3Dx%5E5%7D%5C%5C%5C%5C+%5Cmathsf%7B20-5p%3D5%7D%5C%5C%5C%5C+%5Cmathsf%7B5p%3D20-5%7D%5C%5C%5C%5C+%5Cmathsf%7B5p%3D15%7D%5C%5C%5C%5C+%5Cmathsf%7Bp%3D%5Cdfrac%7B15%7D%7B5%7D%7D)
![\mathsf{p=3} \mathsf{p=3}](https://tex.z-dn.net/?f=%5Cmathsf%7Bp%3D3%7D)
O termo em
ocorre na 4ª posição.
• Termo em![\mathsf{x^{15}}: \mathsf{x^{15}}:](https://tex.z-dn.net/?f=%5Cmathsf%7Bx%5E%7B15%7D%7D%3A)
![\mathsf{x^{20-5p}=x^{15}}\\\\ \mathsf{20-5p=15}\\\\ \mathsf{5p=20-15}\\\\ \mathsf{5p=5}\\\\ \mathsf{p=\dfrac{5}{5}} \mathsf{x^{20-5p}=x^{15}}\\\\ \mathsf{20-5p=15}\\\\ \mathsf{5p=20-15}\\\\ \mathsf{5p=5}\\\\ \mathsf{p=\dfrac{5}{5}}](https://tex.z-dn.net/?f=%5Cmathsf%7Bx%5E%7B20-5p%7D%3Dx%5E%7B15%7D%7D%5C%5C%5C%5C+%5Cmathsf%7B20-5p%3D15%7D%5C%5C%5C%5C+%5Cmathsf%7B5p%3D20-15%7D%5C%5C%5C%5C+%5Cmathsf%7B5p%3D5%7D%5C%5C%5C%5C+%5Cmathsf%7Bp%3D%5Cdfrac%7B5%7D%7B5%7D%7D)
![\mathsf{p=1} \mathsf{p=1}](https://tex.z-dn.net/?f=%5Cmathsf%7Bp%3D1%7D)
O termo em
ocorre na 2ª posição.
—————
O coeficiente de qualquer potência de x no desenvolvimento é
![\mathsf{c_{p+1}=\dbinom{10}{p}\cdot 2^{10-p}\cdot a^p} \mathsf{c_{p+1}=\dbinom{10}{p}\cdot 2^{10-p}\cdot a^p}](https://tex.z-dn.net/?f=%5Cmathsf%7Bc_%7Bp%2B1%7D%3D%5Cdbinom%7B10%7D%7Bp%7D%5Ccdot+2%5E%7B10-p%7D%5Ccdot+a%5Ep%7D)
De acordo com o enunciado, queremos que
![\mathsf{c_4=c_2}\\\\ \mathsf{c_{3+1}=c_{1+1}}\\\\ \mathsf{\dbinom{10}{3}\cdot 2^{10-3}\cdot a^3=\dbinom{10}{1}\cdot 2^{10-1}\cdot a^1}\\\\\\ \mathsf{\dfrac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}\cdot 2^7\cdot a^3=10\cdot 2^9\cdot a} \mathsf{c_4=c_2}\\\\ \mathsf{c_{3+1}=c_{1+1}}\\\\ \mathsf{\dbinom{10}{3}\cdot 2^{10-3}\cdot a^3=\dbinom{10}{1}\cdot 2^{10-1}\cdot a^1}\\\\\\ \mathsf{\dfrac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}\cdot 2^7\cdot a^3=10\cdot 2^9\cdot a}](https://tex.z-dn.net/?f=%5Cmathsf%7Bc_4%3Dc_2%7D%5C%5C%5C%5C+%5Cmathsf%7Bc_%7B3%2B1%7D%3Dc_%7B1%2B1%7D%7D%5C%5C%5C%5C+%5Cmathsf%7B%5Cdbinom%7B10%7D%7B3%7D%5Ccdot+2%5E%7B10-3%7D%5Ccdot+a%5E3%3D%5Cdbinom%7B10%7D%7B1%7D%5Ccdot+2%5E%7B10-1%7D%5Ccdot+a%5E1%7D%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B%5Cdfrac%7B10%5Ccdot+9%5Ccdot+8%7D%7B3%5Ccdot+2%5Ccdot+1%7D%5Ccdot+2%5E7%5Ccdot+a%5E3%3D10%5Ccdot+2%5E9%5Ccdot+a%7D)
Colocando fatores comuns em evidência,
![\mathsf{(10\cdot 2^7\cdot a)\cdot \dfrac{9\cdot 8}{3\cdot 2\cdot 1}\cdot a^2=(10\cdot 2^7\cdot a)\cdot 2^2} \mathsf{(10\cdot 2^7\cdot a)\cdot \dfrac{9\cdot 8}{3\cdot 2\cdot 1}\cdot a^2=(10\cdot 2^7\cdot a)\cdot 2^2}](https://tex.z-dn.net/?f=%5Cmathsf%7B%2810%5Ccdot+2%5E7%5Ccdot+a%29%5Ccdot+%5Cdfrac%7B9%5Ccdot+8%7D%7B3%5Ccdot+2%5Ccdot+1%7D%5Ccdot+a%5E2%3D%2810%5Ccdot+2%5E7%5Ccdot+a%29%5Ccdot+2%5E2%7D)
![\mathsf{(10\cdot 2^7\cdot a)\cdot 12\cdot a^2=(10\cdot 2^7\cdot a)\cdot 4}\\\\ \mathsf{(10\cdot 2^7\cdot a)\cdot 12\cdot a^2-(10\cdot 2^7\cdot a)\cdot 4=0} \mathsf{(10\cdot 2^7\cdot a)\cdot 12\cdot a^2=(10\cdot 2^7\cdot a)\cdot 4}\\\\ \mathsf{(10\cdot 2^7\cdot a)\cdot 12\cdot a^2-(10\cdot 2^7\cdot a)\cdot 4=0}](https://tex.z-dn.net/?f=%5Cmathsf%7B%2810%5Ccdot+2%5E7%5Ccdot+a%29%5Ccdot+12%5Ccdot+a%5E2%3D%2810%5Ccdot+2%5E7%5Ccdot+a%29%5Ccdot+4%7D%5C%5C%5C%5C+%5Cmathsf%7B%2810%5Ccdot+2%5E7%5Ccdot+a%29%5Ccdot+12%5Ccdot+a%5E2-%2810%5Ccdot+2%5E7%5Ccdot+a%29%5Ccdot+4%3D0%7D)
Fatorando por agrupamento,
![\mathsf{(10\cdot 2^7\cdot a)\cdot (12a^2-4)=0}\\\\ \mathsf{4\cdot 10\cdot 2^7 \cdot a\cdot (3a^2-1)=0}\\\\ \mathsf{a\cdot (3a^2-1)=0} \mathsf{(10\cdot 2^7\cdot a)\cdot (12a^2-4)=0}\\\\ \mathsf{4\cdot 10\cdot 2^7 \cdot a\cdot (3a^2-1)=0}\\\\ \mathsf{a\cdot (3a^2-1)=0}](https://tex.z-dn.net/?f=%5Cmathsf%7B%2810%5Ccdot+2%5E7%5Ccdot+a%29%5Ccdot+%2812a%5E2-4%29%3D0%7D%5C%5C%5C%5C+%5Cmathsf%7B4%5Ccdot+10%5Ccdot+2%5E7+%5Ccdot+a%5Ccdot+%283a%5E2-1%29%3D0%7D%5C%5C%5C%5C+%5Cmathsf%7Ba%5Ccdot+%283a%5E2-1%29%3D0%7D)
![\begin{array}{rcl} \mathsf{a=0}&~\textsf{ ou }~&\mathsf{3a^2-1=0}\\\\ \mathsf{a=0}&~\textsf{ ou }~&\mathsf{3a^2=1}\\\\ \mathsf{a=0}&~\textsf{ ou }~&\mathsf{a^2=\dfrac{\,1\,}{3}}\\\\ \mathsf{a=0}&~\textsf{ ou }~&\mathsf{a=\pm\,\dfrac{1}{\sqrt{3}}} \end{array} \begin{array}{rcl} \mathsf{a=0}&~\textsf{ ou }~&\mathsf{3a^2-1=0}\\\\ \mathsf{a=0}&~\textsf{ ou }~&\mathsf{3a^2=1}\\\\ \mathsf{a=0}&~\textsf{ ou }~&\mathsf{a^2=\dfrac{\,1\,}{3}}\\\\ \mathsf{a=0}&~\textsf{ ou }~&\mathsf{a=\pm\,\dfrac{1}{\sqrt{3}}} \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D+%5Cmathsf%7Ba%3D0%7D%26amp%3B%7E%5Ctextsf%7B+ou+%7D%7E%26amp%3B%5Cmathsf%7B3a%5E2-1%3D0%7D%5C%5C%5C%5C+%5Cmathsf%7Ba%3D0%7D%26amp%3B%7E%5Ctextsf%7B+ou+%7D%7E%26amp%3B%5Cmathsf%7B3a%5E2%3D1%7D%5C%5C%5C%5C+%5Cmathsf%7Ba%3D0%7D%26amp%3B%7E%5Ctextsf%7B+ou+%7D%7E%26amp%3B%5Cmathsf%7Ba%5E2%3D%5Cdfrac%7B%5C%2C1%5C%2C%7D%7B3%7D%7D%5C%5C%5C%5C+%5Cmathsf%7Ba%3D0%7D%26amp%3B%7E%5Ctextsf%7B+ou+%7D%7E%26amp%3B%5Cmathsf%7Ba%3D%5Cpm%5C%2C%5Cdfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%7D+%5Cend%7Barray%7D)
a = 0 não se aplica, pois nesse caso não teríamos um binômio de fato.
Logo, temos dois valores possíveis para a:
<——— esta é a resposta.
Bons estudos! :-)
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Encontrar o valor de a, de modo que os coeficientes do termo em
——————————
• Encontrar a fórmula do termo geral do binômio.
No desenvolvimento de
o termo da posição p + 1 é dado por
com 0 ≤ p ≤ n.
Para o binômio dado,
temos
A fórmula do termo geral para este binômio é
Perceba que o expoente de x na posição p + 1 é sempre dado por 20 – 5p.
Sendo assim, vamos encontrar em que posições ocorrem os termos em
• Termo em
O termo em
• Termo em
O termo em
—————
O coeficiente de qualquer potência de x no desenvolvimento é
De acordo com o enunciado, queremos que
Colocando fatores comuns em evidência,
Fatorando por agrupamento,
a = 0 não se aplica, pois nesse caso não teríamos um binômio de fato.
Logo, temos dois valores possíveis para a:
Bons estudos! :-)
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