Question

Escreva uma lei fechada para a seguinte soma

     S(n) = 1³ + 4³ + 7³ + 10³ + ... + (3n − 2)³,     n ∈ ℕ*

e expresse a sua resposta em termos de n.

Answer 1

$\mathsf{S(n)=1^{3}+4^{3}+7^{3}+...+(3n-2)^{3}=\displaystyle\sum\limits_{k=1}^{n}(3k-2)^{3}}$

Sabemos simplificar somas telescópicas, da forma

$\mathsf{\displaystyle\sum_{k=x_{0}}^{x}\big[a_{k+1}-a_{k}\big]=\sum_{k=x_{0}}^{x}\Delta(a_{k})}$

Por cancelamento de termos, obtemos

$\mathsf{\displaystyle\sum_{k=x_{0}}^{x}\Delta(a_{k})=a_{x+1}-a_{x_{0}}}$
________________________________

Podemos perceber que a primeira diferença positiva (operador definido acima por $\Delta$) de polinômios de grau j é um polinômio de grau j - 1:

$\bullet\,\,\mathsf{\Delta(k^{4})=(k+1)^{4}-k^{4}=\big(k^{4}+4k^{3}+6k^{2}+4k+1\big)-k^{4}=}\\\\\mathsf{=4k^{3}+6k^{2}+4k+1}\\\\\\\bullet\,\,\mathsf{\Delta(k^{3})=(k+1)^{3}-k^{3}=\big(k^{3}+3k^{2}+3k+1\big)-k^{3}=}\\\\\mathsf{=3k^{2}+3k+1}\\\\\\\bullet\,\,\mathsf{\Delta(k^{2})=(k+1)^{2}-k^{2}=\big(k^{2}+2k+1\big)-k^{2}=}\\\\\mathsf{=2k+1}\\\\\\\bullet\,\,\mathsf{\Delta(k)=(k+1)-k=}\\\\\mathsf{=1}$

Sabemos somar os polinômios encontrados, pois são diferenças de termos consecutivos de uma sequência, portanto, a ideia é escrever o somando $\mathsf{(3k - 2)^{3} = 27k^{3} - 54k^{2} + 36k - 8}$ como uma combinação linear dos polinômios $\mathsf{\Delta(k^{4}),\,\Delta(k^{3}),\,\Delta(k^{2})}$ e $\mathsf{\Delta(k)}$


Queremos encontrar a, b, c, d constantes reais que satisfaçam
$\mathsf{a\cdot\Delta(k^{4})+b\cdot\Delta(k^{3})+c\cdot\Delta(k^{2})+d\cdot\Delta(k)=27k^{3}-54k^{2}+36k-8}$

Expandindo o lado esquerdo:

$\mathsf{a\cdot\Delta(k^{4})+b\cdot\Delta(k^{3})+c\cdot\Delta(k^{2})+d\cdot\Delta(k)=}\\\\\mathsf{=a\cdot(4k^{3}+6k^{2}+4k+1)+b\cdot(3k^{2} + 3k + 1)+c\cdot(2k+1)+d\cdot1}\\\\\mathsf{=\big(4ak^{3}+6ak^{2}+4ak+a\big)+\big(3bk^{2}+3bk+b\big)+\big(2ck+c\big)+d}\\\\\mathsf{=(4a)k^{3}+(6a+3b)k^{2}+(4a+3b+2c)k+(a+b+c+d)}$

Dois polinômios são iguais se e somente se os coeficientes de potências de mesmo grau são iguais, portanto

$\mathsf{a\cdot\Delta(k^{4})+b\cdot\Delta(k^{3})+c\cdot\Delta(k^{2})+d\cdot\Delta(k)=27k^{3}-54k^{2}+36k-8}\\\\\Longleftrightarrow\\\\\begin{cases}\mathsf{4a=27}\\\mathsf{6a+3b=-54}\\\mathsf{4a+3b+2c=36}\\\mathsf{a+b+c+d=-8}\end{cases}$

Esse sistema pode ser resolvido de cima para baixo:

$\bullet\,\,\mathsf{4a=27\,\,\,\,\Longleftrightarrow\,\,\,\,\boxed{\mathsf{a=\frac{27}{4}}}}\\\\\\\bullet\,\,\mathsf{6a+3b=-54\,\,\Leftrightarrow\,\,b=-18-2a=-18-2\big(\frac{27}{4}\big)=-\frac{18\cdot2}{2}-\frac{27}{2}}\\\\\Longleftrightarrow\,\,\,\,\boxed{\mathsf{b=-\frac{63}{2}}}\\\\\\\bullet\,\,\mathsf{4a+3b+2c=36\,\,\Leftrightarrow\,\,2c=36-4\big(\frac{27}{4}\big)-3\big(-\frac{63}{2}\big)=36-27+\frac{189}{2}}\\\\\Leftrightarrow\,\,\,\mathsf{2c=9+\frac{189}{2}=\frac{18}{2}+\frac{189}{2}=\frac{207}{2}}$

$\mathsf{\Longleftrightarrow\,\,\,\,\boxed{\mathsf{c=\frac{207}{4}}}}\\\\\\\bullet\,\,\mathsf{a+b+c+d=-8\,\,\Leftrightarrow\,\,d=-8-\big(\frac{27}{4}\big)-\big(-\frac{63}{2}\big)-\big(\frac{207}{4}\big)=}\\\\\mathsf{=-\frac{32}{4}-\frac{27}{4}+\frac{126}{4}-\frac{207}{4}=-\frac{140}{4}\,\,\,\,\Longleftrightarrow\,\,\,\,\boxed{\mathsf{d=-35}}}$

Portanto,

$\boxed{\boxed{\mathsf{(3k-2)^{3}=\frac{27}{4}\Delta(k^{4})-\frac{63}{2}\Delta(k^{3})+\frac{207}{4}\Delta(k^{2})-35\Delta(k)}}}$


Agora, vamos achar$\mathsf{\displaystyle\sum_{k=1}^{n}\Delta(k^{m})\,\,\forall\,\,1\le m\le4}$:

$\bullet\,\,\mathsf{\displaystyle\sum_{k=1}^{n}\Delta(k^{1})=\sum_{k=1}^{n}\Delta(k)=\sum_{k=1}^{n}1=n}\\\\\\\bullet\,\,\mathsf{\displaystyle\sum_{k=1}^{n}\Delta(k^{2})=k^{2}\bigg|_{1}^{n+1}=(n+1)^{2}-1^{2}=n^{2}+2n}\\\\\\\bullet\,\,\mathsf{\displaystyle\sum_{k=1}^{n}\Delta(k^{3})=k^{3}\bigg|_{1}^{n+1}=(n+1)^{3}-1^{3}=n^{3}+3n^{2}+3n}\\\\\\\bullet\,\,\mathsf{\displaystyle\sum_{k=1}^{n}\Delta(k^{4})=k^{4}\bigg|_{1}^{n+1}=(n+1)^{4}-1^{4}=n^{4}+4n^{3}+6n^{2}+4n}$

Então:

$\mathsf{S(n)=\sum\limits_{k=1}^{n}(3k-2)^{3}=\sum\limits_{k=1}^{n}\big[\frac{27}{4}\Delta(k^{4})-\frac{63}{2}\Delta(k^{3})+\frac{207}{4}\Delta(k^{2})-35\Delta(k)\big]}\\\\\\\mathsf{=\frac{27}{4}\sum\limits_{k=1}^{n}\Delta(k^{4})-\frac{63}{2}\sum\limits_{k=1}^{n}\Delta(k^{3})+\frac{207}{4}\sum\limits_{k=1}^{n}\Delta(k^{2})-35\sum\limits_{k=1}^{n}\Delta(k)}\\\\\\=\mathsf{\frac{27}{4}\big(n^{4}+4n^{3}+6n^{2}+4n\big)-\frac{63}{2}\big(n^{3}+3n^{2}+3n\big)+\frac{207}{4}\big(n^{2}+2n\big)-35n}$

$\mathsf{=\frac{27}{4}n^{4}+27n^{3}+\frac{81}{2}n^{2}+27n-\frac{63}{2}n^{3}-\frac{189}{2}n^{2}-\frac{189}{2}n+\frac{207}{4}n^{2}+\frac{207}{2}n-35n}\\\\\\\mathsf{=\frac{27}{4}n^{4}+\big(27-\frac{63}{2}\big)n^{3}+\big(\frac{81}{2}-\frac{189}{2}+\frac{207}{4}\big)n^{2}+\big(27-\frac{189}{2}+\frac{207}{2}-35\big)n}\\\\\\\mathsf{=\frac{27}{4}n^{4}-\frac{9}{2}n^{3}-\frac{9}{4}n^{2}+n}\\\\\\\mathsf{=\frac{1}{4}\big(27n^{4}-18n^{3}-9n^{2}+4n\big)}$

Essa é a solução.

$\boxed{\boxed{\mathsf{\sum_{k=1}^{n}(3k-2)^{3}=\frac{1}{4}\big(27n^{4}-18n^{3}-9n^{2}+4n\big)}}}$