Question

Compute the sum:


`\mathsf{\displaystyle\sum_(k=0)^n}` (1 + k²) · k!

Express your answer in terms of n.

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Answer 1

You could use perturbation method to calculate this sum. Let's start from:


`S_n=\sum\limits_(k=0)^nk!\\\\\\\(1)\qquad\boxed{S_(n+1)=S_n+(n+1)!}`

On the other hand, we have:


`S_(n+1)=\sum\limits_(k=0)^(n+1)k!=0!+\sum\limits_(k=1)^(n+1)k!=1+\sum\limits_(k=1)^(n+1)k!=1+\sum\limits_(k=0)^(n)(k+1)!=\\\\\\=1+\sum\limits_(k=0)^(n)k!(k+1)=1+\sum\limits_(k=0)^(n)(k\cdot k!+k!)=1+\sum\limits_(k=0)^(n)k\cdot k!+\sum\limits_(k=0)^(n)k!\\\\\\(2)\qquad \boxed{S_(n+1)=1+\sum\limits_(k=0)^(n)k\cdot k!+S_n}`

So from (1) and (2) we have:


`\begin{cases}S_(n+1)=S_n+(n+1)!\\\\S_(n+1)=1+\sum\limits_(k=0)^(n)k\cdot k!+S_n\end{cases}\\\\\\ S_n+(n+1)!=1+\sum\limits_(k=0)^(n)k\cdot k!+S_n\\\\\\ (\star)\qquad\boxed{\sum\limits_(k=0)^(n)k\cdot k!=(n+1)!-1}`

Now, let's try to calculate sum
`\sum\limits_(k=0)^(n)k\cdot k!`, but this time we use perturbation method.


`S_n=\sum\limits_(k=0)^nk\cdot k!\\\\\\ \boxed{S_(n+1)=S_n+(n+1)(n+1)!}\\\\\\`

but:


`S_(n+1)=\sum\limits_(k=0)^(n+1)k\cdot k!=0\cdot0!+\sum\limits_(k=1)^(n+1)k\cdot k!=0+\sum\limits_(k=0)^(n)(k+1)(k+1)!=\\\\\\= \sum\limits_(k=0)^(n)(k+1)(k+1)k!=\sum\limits_(k=0)^(n)(k^2+2k+1)k!=\\\\\\= \sum\limits_(k=0)^(n)\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_(k=0)^(n)(k^2+1)k!+\sum\limits_(k=0)^n2k\cdot k!=\\\\\\=\sum\limits_(k=0)^(n)(k^2+1)k!+2\sum\limits_(k=0)^nk\cdot k!=\sum\limits_(k=0)^(n)(k^2+1)k!+2S_n\\\\\\ \boxed{S_(n+1)=\sum\limits_(k=0)^(n)(k^2+1)k!+2S_n}`

When we join both equation there will be:


`\begin{cases}S_(n+1)=S_n+(n+1)(n+1)!\\\\S_(n+1)=\sum\limits_(k=0)^(n)(k^2+1)k!+2S_n\end{cases}\\\\\\ S_n+(n+1)(n+1)!=\sum\limits_(k=0)^(n)(k^2+1)k!+2S_n\\\\\\\\ \sum\limits_(k=0)^(n)(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n=\\\\\\= (n+1)(n+1)!-\sum\limits_(k=0)^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]=\\\\\\=(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1=\\\\\\= n(n+1)!+1`

So the answer is:


`\boxed{\sum\limits_(k=0)^(n)(1+k^2)k!=n(n+1)!+1}`

Sorry for my bad english, but i hope it won't be a big problem :)