Question

Mostre que:

$\displaystyle\sum_{k\:\!=\:\!1}^{n}k^3=1^3+2^3+3^3+\cdots+n^3=\left[\dfrac{n(n+1)}{2}\right]^2$

Answer 1

Em primeiro lugar, lembremo-nos que k³ pode ser reescrito como:

$\sf k^3=k^3+0+0\\\\ k^3=k^3+3k^2-3\!\:\!\:\!k^2+2\!\:\!\:\!k-2k\\\\ k^3=k^3+3k^2+2\!\:\!\:\!k-3\!\:\!\:\!k^2-2k\\\\ k^3=k^3+3k^2+2\!\:\!\:\!k-3\!\:\!\:\!k^2-3k+k\\\\ k^3=k(k^2+3\!\:\!\:\!k+2)-3k\!\:\!\:\!(k+1)+k\\\\ k^3=k(k^2+k+2k+2)-3\!\:\!\:\!k(k+1)+k\\\\ k^3=k\big[k(k+1)+2(k+1)\big]-3\!\:\!\:\!k(k+1)+k\\\\ k^3=k\big[(k+1)(k+2)\big]-3k(k+1)+k\\\\ k^3=k(k+1)(k+2)-3k(k+1)+k\qquad (\: i\: )$

Em seguida, tomando o somatório de 1 a n em ambos os membros da equação ( i ), ficaremos com:

$\!\sf\displaystyle\sum_{k\:\!=\:\!1}^{n}k^3=\sum_{k\:\!=\:\!1}^{n}\big[k(k+1)(k+2)-3\!\:\!\:\!k(k+1)+k\big]\\\\\\ \sum_{k\:\!=\:\!1}^{n}k^3=\sum_{k\:\!=\:\!1}^{n}k(k+1)(k+2)-\sum_{k\:\!\:\!=\ \!\!1}^{n}3k(k+1)+\sum_{k\:\!=\:\!1}^{n}k\\\\\\ \sum_{k\:\!=\:\!1}^{n}k^3=\sum_{k\:\!=\:\!1}^{n}k(k+1)(k+2)-3\cdot\!\sum_{k\:\!=\:\!1}^{n}k(k+1)+\sum_{k\:\!=1}^{n}k\qquad(\:ii\:)$

Da Teoria dos Números, sabemos que, para todo k inteiro positivo, temos sempre:

$\sf k=1!\dbinom{k}{1}\\ \\ k(k+1)=2!\dbinom{k+1}{2}\\\\ k(k+1)(k+2)=3!\dbinom{k+2}{3}\\\\ k(k+1)(k+2)(k+3)=4!\dbinom{k+3}{4}\\ \vdots\\ k(k+1)(k+2)\cdots (k+n-1)=n!\dbinom{k+n-1}{n}$

, em que n ≥ 1 é justamente a quantidade de inteiros consecutivos a serem multiplicados (número exato de fatores no lado esquerdo da igualdade). Dessa forma, a equação ( ii ) tornar-se-á:

$\!\sf \displaystyle\sum_{k\:\!=\:\!1}^{n}k^3=\sum_{k\:\!=\:\!1}^{n}\:\!3!\dbinom{k+2}{3}-3\cdot\! \sum_{k\:\!\:\!=\:\!1}^{n}\:\!2!\dbinom{k+1}{2}+\sum_{k\:\!=\:\!1}^{n}\:\!1!\dbinom{k}{1}\\\\\\ \sum_{k\:\!=\:\!1}^{n}k^3=3!\cdot \!\sum_{k\:\!=\:\!1}^{n}\dbinom{k+2}{3}-3\cdot 2!\cdot\!\sum_{k\:\!=\:\!1}^{n}\dbinom{k+1}{2}+1!\cdot\!\sum_{k\:\!=\:\!1}^{n}\dbinom{k}{1}$

 $\vdots$

$\!\sf\displaystyle\sum_{k\:\!=\:\!1}^{n}k^3=3\cdot2\cdot1\cdot \!\sum_{k\:\!\:\!=\:\!1}^{n}\dbinom{k+2}{3}-3\cdot2\cdot1\cdot\!\sum_{k\:\!=\:\!1}^{n}\dbinom{k+1}{2}+\sum_{k\:\!=\:\!1}^{n}\dbinom{k}{1}\\\\\\ \sum_{k\:\!=\:\!1}^{n}k^3=6\cdot\!\sum_{k\:\!\:\!=\:\!1}^{n}\dbinom{k+2}{3}-6\cdot\!\sum_{k\:\!=\:\!1}^{n}\dbinom{k+1}{2}+\sum_{k\:\!=\:\!1}^{n}\dbinom{k}{1}\qquad(\:iii\:)$

Agora, com o auxílio de uma das famosas propriedades do Triângulo de Pascal, que é o Teorema das Colunas (para uma melhor visualização desta propriedade, utilize o triângulo assimétrico da imagem anexada), asseguramos que:

$\!\sf \displaystyle\sum_{k\:\!=\:\!1}^n\dbinom{k+2}{3}=\dbinom{3}{3}+\dbinom{4}{3}+\cdots+\dbinom{n+2}{3}=\dbinom{(n+2)+1}{3+1}=\dbinom{n+3}{4}\\\\\\ \displaystyle\sum_{k\:\!=\:\!1}^n\dbinom{k+1}{2}=\dbinom{2}{2}+\dbinom{3}{2}+\cdots+\dbinom{n+1}{2}=\dbinom{(n+1)+1}{2+1}=\dbinom{n+2}{3}\\\\\\ \displaystyle\sum_{k\:\!=\:\!1}^n\dbinom{k}{1}=\dbinom{1}{1}+\dbinom{2}{1}+\cdots+\dbinom{n}{1}=\dbinom{n+1}{1+1}=\dbinom{n+1}{2}$

Por último, substituindo em ( iii ) cada somatório pelo respectivo número binomial resultante (calculado acima), obteremos:

$\sf \!\displaystyle\sum_{k\:\!=\:\!1}^{n}k^3=6\cdot\dbinom{n+3}{4}-\:\!6\cdot\dbinom{n+2}{3}+\dbinom{n+1}{2}\\\\\\ \:\!\sum_{k\:\!=\:\!1}^{n}k^3=6\cdot \dfrac{(n+3)!}{4!(n+3-4)!}-6\cdot \dfrac{(n+2)!}{3!(n+2-3)!}+\dfrac{(n+1)!}{2!(n+1-2)!}}\\\\\\ \sum_{k\:\!=\:\!1}^{n}k^3=6\cdot \dfrac{(n+3)!}{4!(n-1)!}-6\cdot \dfrac{(n+2)!}{3!(n-1)!}+\dfrac{(n+1)!}{2!(n-1)!}$

 $\vdots$

$\sf \!\displaystyle\sum_{k\:\!=\:\11}^{n}k^3=\diagup\!\!\!\!6\cdot\dfrac{(n+3)(n+2)(n+1)n}{\diagup\!\!\!\!6\cdot4}-\diagup\!\!\!\!6\cdot\dfrac{(n+2)(n+1)n}{\diagup\!\!\!\!6}+\dfrac{(n+1)n}{2}\\\\\\ \sum_{k\:\!=\:\11}^{n}k^3=\dfrac{n(n+1)(n+2)(n+3)}{4}-\dfrac{4}{4}\cdot n(n+1)(n+2)+\dfrac{2}{2}\cdot \dfrac{n(n+1)}{2}\\\\\\ \sum_{k\:\!=\:\11}^{n}k^3=\dfrac{n(n+1)(n+2)(n+3)}{4}-\dfrac{4n(n+1)(n+2)}{4}+\dfrac{2\:\!n(n+1)}{4}\\\\\\ \sum_{k\:\!=\:\11}^{n}k^3=\dfrac{n(n+1)(n+2)(n+3)-4n(n+1)(n+2)+2\:\!n(n+1)}{4}$

 $\vdots$

$\!\sf\displaystyle\sum_{k\:\!=\:\!1}^{n}k^3=\dfrac{n(n+1)\!\:\!\big[(n+2)(n+3)-4(n+2)+2\big]}{4}\\\\\\ \sum_{k\:\!=\:\!1}^{n}k^3=\dfrac{n(n+1)\!\:\!\big[(n+2)(n+3-4)+2\big]}{4}\\\\\\ \sum_{k\:\!=\:\!1}^{n}k^3=\dfrac{n(n+1)\!\:\!\big[(n+2)(n-1)+2\big]}{4}\\\\\\ \sum_{k\:\!=\:\!1}^{n}k^3=\dfrac{n(n+1)(n^2-n+2\:\!n-2+2)}{4}\\\\\\ \sum_{k\:\!=\:\!1}^{n}k^3=\dfrac{n(n+1)(n^2+n)}{4}\\\\\\ \sum_{k\:\!=\:\!1}^{n}k^3=\dfrac{n(n+1)\!\:\!\big[n(n+1)\big]}{4}$

 $\vdots$

$\!\sf\displaystyle\sum_{k\:\!=\:\!1}^{n}k^3=\dfrac{n(n+1)n(n+1)}{2\cdot2}\\\\\\ \sum_{k\:\!=\:\!1}^{n}k^3=\dfrac{n^2(n+1)^2}{2^{\:\!2}}\\\\\\ \boxed{\sf \sum_{k\:\!=\:\!1}^{n}k^3=\bigg[\dfrac{n(n+1)}{2}\bigg]^{\!\:\!2}}$

1.ª Obs.:

$\sf\displaystyle\sum_{k\:\!=\:\!1}^{n}k=1+2+3+\cdots+n=\dfrac{n(n+1)}{2}=\sqrt{\sum_{k\:\!=\:\!1}^{n}k^3$

2.ª Obs.: mesmo sabendo que existem outras formas (mais fáceis que esta) de mostrar isso, optei por esta demonstração, pois, ao meu ver, é de longe a mais interessante (e talvez a mais bela rs).