Question

Sabendo que

$\mathsf{\displaystyle\sum_{k=0}^n}$ k! = S,

e que

(k + 1)! – k! = k · k!,

escreva uma expressão para o somatório

$\mathsf{\displaystyle\sum_{k=0}^n}$ k² · k!

em função de S.

Answer 1

Vamos encontrar $\Delta(a_{k})$, onde $a_{k}=k\cdot k!$:

$\Delta(a_{k})=a_{k+1}-a_{k}\\\\\Delta(a_{k})=(k+1)\cdot(k+1)!-k\cdot k!\\\\\Delta(a_{k})=(k+1)\cdot(k+1)\cdot k!-k\cdot k!\\\\\Delta(a_{k})=(k+1)^{2}\cdot k!-k\cdot k!\\\\\Delta(a_{k})=(k^{2}+2k+1)\cdot k!-k\cdot k!$

Colocando $k!$ em evidência:

$\Delta(a_{k})=(k^{2}+2k+1-k)\cdot k!\\\\\Delta(a_{k})=(k^{2}+k+1)\cdot k!\\\\\Delta(a_{k})=k^{2}\cdot k!+k\cdot k!+k!\\\\\Delta(a_{k})=k^{2}\cdot k!+\Delta(b_{k})+k!$

onde $b_{k}=k!$.

Isolando $k^{2}\cdot k!$:

$k^{2}\cdot k!=\Delta(a_{k})-\Delta(b_{k})-k!\\\\k^{2}\cdot k!=\Delta(a_{k}-b_{k})-k!$

Pois o operador $\Delta$ é linear. Então:

$\displaystyle\sum_{k=0}^{n}k^{2}\cdot k!=\sum_{k=0}^{n}\big[\Delta(a_{k}-b_{k})-k!\big]\\\\\\\sum_{k=0}^{n}k^{2}\cdot k!=\sum_{k=0}^{n}\Delta(a_{k}-b_{k})-\sum_{k=0}^{n}k!\\\\\\\sum_{k=0}^{n}k^{2}\cdot k!=\sum_{k=0}^{n}\Delta(a_{k}-b_{k})-\mathsf{S}$

Por propriedade, temos $\sum\limits_{k=\ell}^{n}\Delta(c_{k})=c_{n+1}-c_{\ell}$, então

$\displaystyle\sum_{k=0}^{n}k^{2}\cdot k!=\big[a_{n+1}-b_{n+1}\big]-\big[a_{0}-b_{0}\big]-\mathsf{S}\\\\\\\sum_{k=0}^{n}k^{2}\cdot k!=\big[a_{n+1}-b_{n+1}\big]-\big[0\cdot0!-0!]-\mathsf{S}\\\\\\\sum_{k=0}^{n}k^{2}\cdot k!=(n+1)\cdot(n+1)!-(n+1)!-\big[0-1\big]-\mathsf{S}\\\\\\\sum_{k=0}^{n}k^{2}\cdot k!=(n+1)!\cdot\big[(n+1)-1\big]+1-\mathsf{S}\\\\\\\boxed{\boxed{\sum_{k=0}^{n}k^{2}\cdot k!=n\cdot(n+1)!+1-\mathsf{S}}}$