Question

(50 PONTOS) a) Obtenha a forma fechada para a soma
$\displaystyle\sum\limits_{n=0}^{k}{(3n^{2}+3n+1)}$

Sugestão: Utilize a sequência $a_{n}=n^{3}$ e tome a diferença entre dois termos consecutivos para obter uma soma telescópica.


b) Com o resultado acima, e sabendo que

$\displaystyle\sum\limits_{n=0}^{k}{1}=k+1\;\;\text{ e }\;\;\displaystyle\sum\limits_{n=0}^{k}{n}=\dfrac{(k+1)\cdot k}{2}$


mostre que a soma dos quadrados dos naturais é dada por

$\displaystyle\sum\limits_{n=0}^{k}{n^{2}}=\dfrac{k\cdot (k+1)\cdot (2k+1)}{6}$

Answer 1

Utilizarei a seguinte fatoração:

$a^{3}-b^{3}=(a-b)\cdot(a^{2}+ab+b^{2})$
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Considerando a sequência $a_{n}=n^{3}$ e avaliando a diferença de dois termos consecutivos, temos:

$(n+1)^{3}-n^{3}=(n+1-n)\cdot([n+1]^{2}+[n+1]\cdot n+n^{2})\\\\(n+1)^{3}-n^{3}=n^{2}+2n+1+n^{2}+n+n^{2}\\\\(n+1)^{3}-n^{3}=3n^{2}+3n+1$

Ou seja

$\displaystyle\sum\limits_{n=0}^{k}3n^{2}+3n+1=\sum\limits_{n=0}^{k}(n+1)^{3}-n^{3}\\\\\\\sum\limits_{n=0}^{k}3n^{2}+3n+1=\sum\limits_{n=0}^{k}(n+1)^{3}-\sum\limits_{n=0}^{k}n^{3}$

Desenvolvendo os somatórios, temos:

$\displaystyle\sum\limits_{n=0}^{k}3n^{2}+3n+1=1^{3}+2^{3}+...+(k+1)^{3}-(0^{3}+1^{3}+2^{3}+...+k^{3})\\\\\\\sum\limits_{n=0}^{k}3n^{2}+3n+1=1^{3}+2^{3}+...+(k+1)^{3}-1^{3}-2^{3}-...-k^{3}\\\\\\\sum\limits_{n=0}^{k}3n^{2}+3n+1=(1^{3}-1^{3})+(2^{3}-2^{3})+...+(k^{3}-k^{3})+(k+1)^{3}\\\\\\\sum\limits_{n=0}^{k}3n^{2}+3n+1=0+0+0+...+0+(k+1)^{3}\\\\\\\boxed{\boxed{\sum\limits_{n=0}^{k}3n^{2}+3n+1=(k+1)^{3}}}$
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b)

Note que

$\displaystyle\sum\limits_{n=0}^{k}3n^{2}+3n+1=\sum\limits_{n=0}^{k}3n^{2}+\sum\limits_{n=0}^{k}3n+\sum\limits_{n=0}^{k}1=(k+1)^{3}$

Tirando as constantes dos somatórios, temos

$\displaystyle3\sum\limits_{n=0}^{k}n^{2}+3\sum\limits_{n=0}^{k}n+\sum\limits_{n=0}^{k}1=(k+1)^{3}$

Usando o que foi dado:

$\displaystyle3\sum\limits_{n=0}^{k}n^{2}+3\dfrac{(k+1)\cdot k}{2}+(k+1)=(k+1)^{3}\\\\\\3\sum\limits_{n=0}^{k}n^{2}=(k+1)^{3}-\dfrac{3}{2}k(k+1)-(k+1)$

Colocando (k + 1) em evidência:

$\displaystyle3\sum\limits_{n=0}^{k}n^{2}=(k+1)\cdot\left((k+1)^{2}-\dfrac{3k}{2}-1\right)\\\\\\3\sum\limits_{n=0}^{k}n^{2}=(k+1)\cdot\left(k^{2}+2k+1-\dfrac{3k}{2}-1\right)\\\\\\3\sum\limits_{n=0}^{k}n^{2}=(k+1)\cdot\left(k^{2}+2k-\dfrac{3k}{2}\right)\\\\\\3\sum\limits_{n=0}^{k}n^{2}=(k+1)\cdot\left(\dfrac{2k^{2}+4k-3k}{2}\right)\\\\\\3\sum\limits_{n=0}^{k}n^{2}=(k+1)\cdot\left(\dfrac{2k^{2}+k}{2}\right)$

Colocando k em evidência:

$\displaystyle3\sum\limits_{n=0}^{k}n^{2}=(k+1)\cdot\dfrac{k\cdot(2k+1)}{2}\\\\\\3\sum\limits_{n=0}^{k}n^{2}=\dfrac{k\cdot(k+1)\cdot(2k+1)}{2}$

Dividindo os dois lados por três:

$\displaystyle\dfrac{3}{3}\sum\limits_{n=0}^{k}n^{2}=\dfrac{k\cdot(k+1)\cdot(2k+1)}{3\cdot2}\\\\\\\boxed{\boxed{\sum\limits_{n=0}^{k}n^{2}=\dfrac{k\cdot(k+1)\cdot(2k+1)}{6}}}$