- #1
DKATyler
- 4
- 0
[SOLVED] Seperating a Summation problem.
The Problem:
Separate a sum into 2 pieces (part of a proof problem).
Using: [tex]X=
\sum^{n}_{k=1}\frac{n!}{(n-k)!}
[/tex]
Solve in relation to n and X:
[tex]
\sum^{n+1}_{k=1}\frac{(n+1)!}{(n+1-k)!}
[/tex]
?
[tex]
\sum^{n}_{k=1}[\frac{(n+1)!}{(n+1-k)!}]+\frac{(n+1)!}{(n+1-[n+1])!}
[/tex]
[tex]
\sum^{n}_{k=1}[\frac{(n)!}{(n-k)!}*\frac{(n+1)}{(n+1-k)}]+\frac{(n+1)!}{(n+1-[n+1])!}
[/tex]
[tex]
(n+1)*\sum^{n}_{k=1}[\frac{(n)!}{(n-k)!}*\frac{1}{(n+1-k)}]+(n+1)!}
[/tex]
I think this is fairly close but, I have no way of getting rid of the 1/(n+1-k) term.
Homework Statement
The Problem:
Separate a sum into 2 pieces (part of a proof problem).
Using: [tex]X=
\sum^{n}_{k=1}\frac{n!}{(n-k)!}
[/tex]
Solve in relation to n and X:
[tex]
\sum^{n+1}_{k=1}\frac{(n+1)!}{(n+1-k)!}
[/tex]
Homework Equations
?
The Attempt at a Solution
[tex]
\sum^{n}_{k=1}[\frac{(n+1)!}{(n+1-k)!}]+\frac{(n+1)!}{(n+1-[n+1])!}
[/tex]
[tex]
\sum^{n}_{k=1}[\frac{(n)!}{(n-k)!}*\frac{(n+1)}{(n+1-k)}]+\frac{(n+1)!}{(n+1-[n+1])!}
[/tex]
[tex]
(n+1)*\sum^{n}_{k=1}[\frac{(n)!}{(n-k)!}*\frac{1}{(n+1-k)}]+(n+1)!}
[/tex]
I think this is fairly close but, I have no way of getting rid of the 1/(n+1-k) term.