Summation of Series: Faulhaber's FormulaDate: 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/ |
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