Help! What are the patterns

I have a tricky question here....

Please just feed in any comments, anything you can spot that I can't

I have been asked to compare the formulae of Sygma (k=1 until n value) of k, Sygma (k=1 until n value) of k(k+1) and Sygma (k=1 until n value) of k(k+1)(k+2)?

Hence comparing these...

Sygma (k=1 until n value) of k = 1/2 n(n+1)
Sygma (k=1 until n value) of k(k+1) = 1/3 n(n+1)(n+2)
Sygma (k=1 until n value) of k(k+1)(k+2) = 1/4 n(n+1)(n+2)(n+3)

I have been asked to comment on these as much as I possibly can??? :confused: Can anyone help?
 
Last edited:
poo, no one has a clue :uhh:
 

benorin

Homework Helper
1,057
7
Here you go, Natasha.

Natasha1 said:
I have a tricky question here....
Please just feed in any comments, anything you can spot that I can't

I have been asked to compare the formulae of Sygma (k=1 until n value) of k, Sygma (k=1 until n value) of k(k+1) and Sygma (k=1 until n value) of k(k+1)(k+2)?
Hence comparing these...
Sygma (k=1 until n value) of k = 1/2 n(n+1)
Sygma (k=1 until n value) of k(k+1) = 1/3 n(n+1)(n+2)
Sygma (k=1 until n value) of k(k+1)(k+2) = 1/4 n(n+1)(n+2)(n+3)
I have been asked to comment on these as much as I possibly can??? :confused: Can anyone help?
Here is a helpful observation: [tex]k(k+1)\cdot\cdot\cdot(k+j)=\frac{(k+j)!}{(k-1)!}[/tex] and [tex]\sum_{k=1}^{n}\frac{(k+j)!}{(k-1)!}=\frac{n(n+j+1)!}{(j+2)n!}[/tex]
Note that [tex]k[/tex], [tex]k(k+1)[/tex], and [tex]k(k+1)(k+2)[/tex] are given by [tex]\frac{(k+j)!}{(k-1)!}[/tex] for [tex]j=0,1,\mbox{ and }2,[/tex] respectively, so that

[tex]\sum_{k=1}^{n} k = \frac{n(n+0+1)!}{(0+2)n!} = \frac{n(n+1)}{2}[/tex],

[tex]\sum_{k=1}^{n} k(k+1) = \frac{n(n+1+1)!}{(1+2)n!} = \frac{n(n+1)(n+2)}{3}[/tex],

and

[tex]\sum_{k=1}^{n} k(k+1)(k+2) = \frac{n(n+2+1)!}{(2+2)n!} = \frac{n(n+1)(n+2)(n+3)}{4}[/tex].
-Ben
 
Last edited:

Tide

Science Advisor
Homework Helper
3,078
0
Ben,

That is REALLY nice!
 
Can anyone see anything else please? :uhh:
 
thanks super Ben!
 

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