Asaleswoman received $40 per day, plus a commission of 2% on all sales over $300 per week. what was her salary for a 5-day week during which her sales amounted to $1260?

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

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

On the other hand, we have:

\begin{gathered}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}\end{gathered}

S

n+1

=

k=0

∑

n+1

k!=0!+

k=1

∑

n+1

k!=1+

k=1

∑

n+1

k!=1+

k=0

∑

n

(k+1)!=

=1+

k=0

∑

n

k!(k+1)=1+

k=0

∑

n

(k⋅k!+k!)=1+

k=0

∑

n

k⋅k!+

k=0

∑

n

k!

(2)

S

n+1

=1+

k=0

∑

n

k⋅k!+S

n

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

\begin{gathered}\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}\end{gathered}

⎩

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎧

S

n+1

=S

n

+(n+1)!

S

n+1

=1+

k=0

∑

n

k⋅k!+S

n

S

n

+(n+1)!=1+

k=0

∑

n

k⋅k!+S

n

(⋆)

k=0

∑

n

k⋅k!=(n+1)!−1

Now, let's try to calculate sum \sum\limits_{k=0}^{n}k\cdot k!

k=0

∑

n

k⋅k! , but this time we use perturbation method.

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

S

n

=

k=0

∑

n

k⋅k!

S

n+1

=S

n

+(n+1)(n+1)!

but:

\begin{gathered}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}\end{gathered}

S

n+1

=

k=0

∑

n+1

k⋅k!=0⋅0!+

k=1

∑

n+1

k⋅k!=0+

k=0

∑

n

(k+1)(k+1)!=

=

k=0

∑

n

(k+1)(k+1)k!=

k=0

∑

n

(k

2

+2k+1)k!=

=

k=0

∑

n

[(k

2

+1)k!+2k⋅k!]=

k=0

∑

n

(k

2

+1)k!+

k=0

∑

n

2k⋅k!=

=

k=0

∑

n

(k

2

+1)k!+2

k=0

∑

n

k⋅k!=

k=0

∑

n

(k

2

+1)k!+2S

n

S

n+1

=

k=0

∑

n

(k

2

+1)k!+2S

n

When we join both equation there will be:

\begin{gathered}\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\end{gathered}

⎩

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⎪

⎪

⎨

⎪

⎪

⎪

⎧

S

n+1

=S

n

+(n+1)(n+1)!

S

n+1

=

k=0

∑

n

(k

2

+1)k!+2S

n

S

n

+(n+1)(n+1)!=

k=0

∑

n

(k

2

+1)k!+2S

n

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)!−

k=0

∑

n

k⋅k!

=

(⋆)

(n+1)(n+1)!−[(n+1)!−1]=

=(n+1)(n+1)!−(n+1)!+1=(n+1)!⋅[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}

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 :)

constants mean the stopped points or the points on a graph that nothing is changing, hope this

step-by-step explanation: