Summation of Series: Faulhaber's Formula

Date: 07/30/2003 at 03:33:28
From: John
Subject: Summation of series

I am asked to solve a series of the form

       n
S(n)= Sum (i*(i+1)*(i+2)*(i+3)*(i+4)), 
      i=1 

e.g.:

S(1) = 1.2.3.4.5
S(2) = 1.2.3.4.5 + 2.3.4.5.6
S(3) = 1.2.3.4.5 + 2.3.4.5.6 + 3.4.5.6.7


I noticed that S(n)= S(n-1)+n*(n+1)*(n+2)*(n+3)*(n+4). But when I 
subtract as in other summations, I end up with the initial series 
again.

S(n) = S(n-1)+n*(n+1)*(n+2)*(n+3)*(n+4)
S(n-1) = S(n-2)+(n-1)*n*(n+1)*(n+2)*(n+3)

S(1) = 1.2.3.4.5
----------------------
S(n)-S(1) = n*(n+1)*(n+2)*(n+3)*(n+4)+.......+2.3.4.5.6

Date: 07/30/2003 at 06:24:44
From: Doctor Jeremiah
Subject: Re: Summation of series

Hi John,

Starting with this:

       n
S(n)= Sum (i*(i+1)*(i+2)*(i+3)*(i+4))
      i=1 

We can multiply the terms together to get:

       n
S(n)= Sum (i^5 + 10i^4 + 35i^3 + 50i^2 + 24i)
      i=1 

Now we can distribute the summation:

        n            n            n            n            n
S(n) = Sum i^5 + 10 Sum i^4 + 35 Sum i^3 + 50 Sum i^2 + 24 Sum i
       i=1          i=1          i=1          i=1          i=1 

Now, there is a way to get the sum of a power. It is called 
Faulhaber's Formula. Here is a link to a page that describes
the general formula and the specific examples we need:

   Eric Weisstein - World of Mathematics
   http://mathworld.wolfram.com/FaulhabersFormula.html 

So our sum becomes:

S(n) = (2n^6+6n^5+5n^4-n^2)/12
     + 10 (6n^5+15n^4+10n^3-n)/30
     + 35 (n^4+2n^3+n^2)/4
     + 50 (2n^3+3n^2+n)/6
     + 24 (n^2+n)/2

Which, after simplifying, becomes:

S(n) = (n^6 + 15n^5 + 85n^4 + 225n^3 + 274n^2 + 120n)/6

Which would give us these answers:

S(n) = (1 + 15 + 85 + 225 + 274 + 120)/6                 = 120
S(n) = (64 + 15*32 + 85*16 + 225*8 + 274*4 + 120*2)/6    = 840
S(n) = (729 + 15*243 + 85*81 + 225*27 + 274*9 + 120*3)/6 = 3360

And they match your answers:

S(1) = 1*2*3*4*5                         = 120
S(2) = 1*2*3*4*5 + 2*3*4*5*6             = 840
S(3) = 1*2*3*4*5 + 2*3*4*5*6 + 3*4*5*6*7 = 3360

Let me know if you need more details on how I got this result.

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/