Factorials Question: how does (n + 16)! - n! equal....

[Trenters4325]

\(\displaystyle (n+16)!-n!\)

\(\displaystyle =n!([n+16][n+15]...[n+1]-1)\)

\(\displaystyle =n*n!+(n+1)*(n+1)!+...+(n+15)(n+15)!\)

How is the third expression equal to the second?

Also, why do my multiplication signs look weird?

 

[jonboy]

In LaTeX do not use * as mulitplication use \(\displaystyle \times\) for muliplication or bullets .

For \(\displaystyle \times\): Type \times enclosed in "tex brackets".
For \(\displaystyle \bullet\): Type \bullet enclosed in "tex brackets".

 

[Trenters4325]

\(\displaystyle (n+16)!-n!\)

\(\displaystyle =n!([n+16][n+15]...[n+1]-1)\)

\(\displaystyle =n[\times]n!+(n+1)[\times](n+1)!+...+(n+15)(n+15)!\)

Okay. Now how do I get rid of the brackets?

 

[jonboy]

\(\displaystyle (n+16)!-n!\)

\(\displaystyle =n!([n+16][n+15]...[n+1]-1)\)

\(\displaystyle =n\times n!+(n+1)\times (n+1)!+...+(n+15)(n+15)!\)

Okay. Now how do I get rid of the brackets?

Im sorry I made myself unclear in my last post. Quote this message to see how I got rid of the brackets.

 

[jonboy]

With bullets to denote multiplication (quote to see how I typed them):

\(\displaystyle (n+16)!-n!\)

\(\displaystyle =n!([n+16][n+15]...[n+1]-1)\)

\(\displaystyle =n\bullet n!+(n+1)\bullet (n+1)!+...+(n+15)(n+15)!\)

 

[pka]

I cannot show you how to deduce line 3 from line 2.
I can show you how line 3 gives line one.

It is well known that for each k,
\(\displaystyle \L
(N + k + 1)! - \left( {N + k } \right)! = \left( {N + k } \right)\left( {N + k } \right)!\).

Now this is line three: \(\displaystyle \L
\begin{array}{rcl}
\sum\limits_{k = 0}^{15} {\left( {N + k} \right)\left( {N + k} \right)!} & = & \sum\limits_{k = 0}^{15} {\left[ {(N + k + 1)! - \left( {N + k} \right)!} \right]} \\
& = & \sum\limits_{k = 0}^{15} {\left( {N + k + 1} \right)! - \sum\limits_{k = 0}^{15} {\left( {N + k} \right)!} } \\
& = & \left( {N + 16} \right)! - N! \\
\end{array}\)

 

[Trenters4325]

\(\displaystyle (n+16)!-n!\)

\(\displaystyle =n!([n+16][n+15]...[n+1]-1)\)

\(\displaystyle =n\times n!+(n+1)\times (n+1)!+...+(n+15)(n+15)!\)

Much better. Now the solution...

 

[pka]

The solution is in reply above your last playing around with LaTeX.
Take time to study its mathematics.
Pretty as it is, LaTeX is not mathematics!

 

[Trenters4325]

That would be a pretty nice solution--if I were going the other way. :shock:

Without knowing the third expression, it would be pretty difficult to think of those steps. So, I still want a forward transition between the second and the third expressions.

 

[pka]

That would be a pretty nice solution--if I were going the other way.

What a ludicrous statement!
Line 1 is equivalent to line 3!
Just turn the argument on its head.
Go from line 1 to line 3, skipping line 2.
Common trick for someone who is working at this level to know!\(\displaystyle \L
\begin{array}{rcl}
\left( {N + 16} \right)! - N! & = & \sum\limits_{k = 0}^{15} {\left( {N + k + 1} \right)! - \sum\limits_{k = 0}^{15} {\left( {N + k} \right)!} } \\
& = & \sum\limits_{k = 0}^{15} {\left[ {(N + k + 1)! - \left( {N + k} \right)!} \right]} \\
& = & \sum\limits_{k = 0}^{15} {\left( {N + k} \right)\left( {N + k} \right)!} \\
\end{array}\)