Exploring Patterns in Sygma Formulas

[Natasha1]

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? Can anyone help?

 

[Natasha1]

poo, no one has a clue :uhh:

 

[benorin]

Here you go, Natasha.

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? Can anyone help?

Here is a helpful observation: $$k(k+1)\cdot\cdot\cdot(k+j)=\frac{(k+j)!}{(k-1)!}$$ and $$\sum_{k=1}^{n}\frac{(k+j)!}{(k-1)!}=\frac{n(n+j+1)!}{(j+2)n!}$$
Note that $$k$$, $$k(k+1)$$, and $$k(k+1)(k+2)$$ are given by $$\frac{(k+j)!}{(k-1)!}$$ for $$j=0,1,\mbox{ and }2,$$ respectively, so that

$$\sum_{k=1}^{n} k = \frac{n(n+0+1)!}{(0+2)n!} = \frac{n(n+1)}{2}$$,

$$\sum_{k=1}^{n} k(k+1) = \frac{n(n+1+1)!}{(1+2)n!} = \frac{n(n+1)(n+2)}{3}$$,

and

$$\sum_{k=1}^{n} k(k+1)(k+2) = \frac{n(n+2+1)!}{(2+2)n!} = \frac{n(n+1)(n+2)(n+3)}{4}$$.
-Ben

 

[Tide]

Ben,

That is REALLY nice!

 

[Natasha1]

Can anyone see anything else please? :uhh:

 

[Natasha1]

thanks super Ben!