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Schach [20]
3 years ago
15

Compute the sum:

Mathematics
1 answer:
Nady [450]3 years ago
8 0
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 :)
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