Mostre que a soma dos n primeiros cubosS(n) = 1^3 + 2^3 + 3^3 + ... + n^3 = pode ser expressa de for.... Pergunta de ideia deLukyo

December 2019 1 230 Report

Mostre que a soma dos n primeiros cubos

S(n) = 1^3 + 2^3 + 3^3 + ... + n^3 = $\mathsf{\displaystyle\sum_{k=1}^n k^3}$

pode ser expressa de forma fechada como

S(n) = [n(n + 1)/2]^2.

Obs.: Não usar indução.

Lista de comentários

superaks

Verified answer

Olá Lukyo.

Teoremas e propriedades usadas.

$\star~\boxed{\boxed{\mathsf{\binom{n}{p}=\dfrac{n!}{p!\cdot(n-p)!}~~~se,~n\ \textgreater \ p}}}~(*)\\\\\\\\\star~\boxed{\boxed{\mathsf{\binom{n}{p}=0~~~se,~p\ \textgreater \ n}}}~(**)\\\\\\\\\star~\boxed{\boxed{\mathsf{\displaystyle\sum _{k=p}^n\binom{k}{p}=\binom{n+1}{p+1}~~~se,~n\ \textgreater \ p}}}~(***)\\\\\\\\\star~\boxed{\boxed{\mathsf{0!=1}}}~(****)$

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- Mostre que a soma dos n primeiros cubos.

$\mathsf{\displaystyle\sum_{k=1}^n k^3=1^3+2^3+3^3+...+n^3}$

Pode ser expressa como.

$\mathsf{\Big[\dfrac{n\cdot(n+1)}{2}\Big]^2}$

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Todo polinômio $\mathsf{f}$ de grau d na variável n, pode ser escrito da seguinte forma.

$\mathsf{f_d(n)=a_0\displaystyle\binom{n}{0}+a_1\displaystyle\binom{n}{1}+a_2\displaystyle\binom{n}{2}+...+a_d\displaystyle\binom{n}{d}}$

Vamos escrever o polinômio k³ no mesmo formato.

$\mathsf{f_3(k)=k^3}\\\\\mathsf{f_3(k)=a_0\displaystyle\binom{k}{0}+a_1\displaystyle\binom{k}{1}+a_2\displaystyle\binom{k}{2}+a_3\displaystyle\binom{k}{3}}$

Para se calcular os coeficientes $\mathsf{a_0,a_1,a_2}$ e $\mathsf{a_3}$ , basta substituir k pelos seus respectivos índices.

$\mathsf{Para~a_0,~k=0}\\\\\mathsf{0^3=a_0\displaystyle\binom{0}{0}+a_1\displaystyle\binom{0}{1}+a_2\displaystyle\binom{0}{2}+a_3\displaystyle\binom{0}{3}}\\\\\\\mathsf{0=a_0\cdot0+a_1\cdot0+a_2\cdot0+a_3\cdot0}\\\\\mathsf{0=0}\\\\\mathsf{a_0=0}\\\\\\\\\mathsf{Para~a_1,~k=1}\\\\\mathsf{1^3=0\displaystyle\binom{1}{0}+a_1\displaystyle\binom{1}{1}+a_2\displaystyle\binom{1}{2}+a_3\displaystyle\binom{1}{3}}\\\\\\\mathsf{1=0+a_1\cdot\dfrac{1!}{1!\cdot(1-1)!}+a_2\cdot0+a_3\cdot0}$

$\mathsf{1=a_1\cdot\dfrac{1}{1\cdot1}}\\\\\\\mathsf{1=a_1}\\\\\\\mathsf{Para~a_2,~k=2}\\\\\\\mathsf{2^3=0\displaystyle\binom{2}{0}+1\displaystyle\binom{2}{1}+a_2\displaystyle\binom{2}{2}+a_3\displaystyle\binom{2}{3}}\\\\\\\mathsf{8=0+1\cdot\dfrac{2!}{1!\cdot(2-1)!}+a_2\cdot\dfrac{2!}{2!\cdot(2-2)!}+a_3\cdot0}\\\\\\\mathsf{8=\dfrac{2}{1\cdot1}+a_2\cdot\dfrac{\diagup\!\!\!\!2}{\diagup\!\!\!\!2\cdot1}}\\\\\\\mathsf{8=2+a_2}\\\\\\\mathsf{8-2=a_2}\\\\\\\mathsf{6=a_2}$

$\mathsf{Para~a_3,~k=3}\\\\\\\mathsf{3^3=0\displaystyle\binom{3}{0}+1\displaystyle\binom{3}{1}+6\displaystyle\binom{3}{2}+a_3\displaystyle\binom{3}{3}}\\\\\\\mathsf{27=0+1\cdot\dfrac{3!}{1!\cdot(3-1)!}+6\cdot\dfrac{3!}{2!\cdot(3-2)!}+a_3\cdot\dfrac{\diagup\!\!\!\!3!}{\diagup\!\!\!\!3!\cdot(3-3)!}}\\\\\\\mathsf{27=\dfrac{3\cdot\diagup\!\!\!\!2!}{1\cdot\diagup\!\!\!\!2!}+6\cdot\dfrac{3\cdot\diagup\!\!\!\!2!}{\diagup\!\!\!\!2!\cdot1!}+a_3\cdot\dfrac{1}{1\cdot1}}\\\\\\\mathsf{27=3+18+a_3}$

$\mathsf{27-21=a_3}\\\\\mathsf{6=a_3}$

Portanto temos.

$\mathsf{k^3=1\displaystyle\binom{k}{1}+6\displaystyle\binom{k}{2}+6\displaystyle\binom{k}{3}}$

Aplicando somatório em cada termo.

$\mathsf{\displaystyle\sum_{k=1}^n k^3=\displaystyle\sum_{k=1}^n1\displaystyle\binom{k}{1}+\displaystyle\sum_{k=1}^n6\displaystyle\binom{k}{2}+\displaystyle\sum_{k=1}^n6\displaystyle\binom{k}{3}}\\\\\\\\\mathsf{\displaystyle\sum_{k=1}^n k^3=1\binom{n+1}{1+1}+6\binom{n+1}{2+1}+6\binom{n+1}{3+1}}~~(***)\\\\\\\\\mathsf{\displaystyle\sum_{k=1}^n k^3=1\cdot\dfrac{(n+1)!}{2!\cdot[(n+1)-2]}+6\cdot\dfrac{(n+1)!}{3!\cdot[(n+1)-3]}+6\cdot\dfrac{(n+1)!}{3!\cdot[(n+1)-4]}}$

$\mathsf{\displaystyle\sum_{k=1}^n k^3=\dfrac{(n+1)!}{2!\cdot(n-1)!}+\diagup\!\!\!\!3!\cdot\dfrac{(n+1)!}{\diagup\!\!\!\!3!\cdot(n-2)!}+\diagup\!\!\!\!3!\cdot\dfrac{(n+1)!}{4\cdot\diagup\!\!\!\!3!\cdot(n-3)!}}\\\\\\\\\mathsf{\displaystyle\sum_{k=1}^n k^3=\dfrac{(n+1)\cdot n\cdot(n-1)!}{2!\!\cdot\!(n-1)!}\!+\!\dfrac{(n+1)\cdot n\!\cdot\!(n-1)\cdot(n-2)!}{(n-2)!}+\dfrac{(n+1)!}{4\cdot(n-3)!}}$

$\mathsf{\displaystyle\sum_{k=1}^n k^3=\dfrac{n\cdot(n+1)}{2}+(n+1)\cdot n\cdot(n-1)+\dfrac{(n+1)\cdot n\cdot(n-1)\cdot(n-2)}{4}}\\\\\\\\\mathsf{\displaystyle\sum_{k=1}^n k^3=(n^2+n)\cdot\Big[\dfrac{1}{2}+(n-1)+\dfrac{(n-1)\cdot(n-2)}{4}\Big]}\\\\\\\\\mathsf{\displaystyle\sum_{k=1}^n k^3=(n^2+n)\cdot\Big[\dfrac{1}{2}\cdot\dfrac{2}{2}+\dfrac{4}{4}\cdot(n-1)+\dfrac{n^2-2n-n+2}{4}\Big]}$

$\mathsf{\displaystyle\sum_{k=1}^n k^3=(n^2+n)\cdot\Big[\dfrac{2}{4}+\dfrac{4n-4}{4}+\dfrac{n^2-3n+2}{4}\Big]}\\\\\\\\\mathsf{\displaystyle\sum_{k=1}^n k^3=(n^2+n)\cdot\Big[\dfrac{n^2+n}{4}\Big]}\\\\\\\\\boxed{\mathsf{\displaystyle\sum_{k=1}^n k^3=\Big[\dfrac{n\cdot(n+1)}{2}\Big]^2}}$

Concluímos o que queríamos mostrar.

Dúvidas? comente.

2 votes Thanks 2

Lukyo Muito bom! Obrigado :)

superaks :D