If S(n,r)=1^r+2^r+... +n^r,
and assuming that you know S(n,1), S(n,2) and S(n,3),
derive the formula for S(n,4) in terms of n.

Start with the fifth powers:

 

1^5 = 1^5

2^5 = (1+1)^5  = 1^5 + 5*1^4*1 + 10*1^3*1^2 + 10*1^2*1^3 + 5*1^1*1^4 + 1^5

 

3^5 = (2+1)^5 = 2^5 + 5*2^4*1 + 10*2^3*1^2 + 10*2^2^1^3 + 5*2^1*1^4 + 1^5

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n^5 = (n-1 + 1)^5 = (n-1)^5 + 5*(n-1)^4*1 + 10*(n-1)^3*1^2 + 10*(n-1)^2*1^3) + 5*(n-1)*1^4 + 1^5

 

Summing all the first and the second members, and simplifying:

 

n^5 = 5*S(n-1)^4 + 10*S(n-1)^3 + 10*S(n-1)^2 + 5*S(n-1) + (n-1)*1

 

S(n-1)^4 = n^5 – 10*S(n-1)^3 – 10*S(n-1)^2 – 5*S(n-1) – (n-1)

 

S(n-1)^3 = S(n)^3 – n^3

S(n-1)^2 = S(n)^2 – n^2

S(n-1) = S(n) – n

 

S(n)^4 = S(n-1)^4 + n^4

 

where S(n) is the sum of the n first natural numbers; S(n)^2 is the sum of the second powers of the first n natural numbers, S(n)^3 is the sum of the third powers of the first n natural numbers, and so on.

 

We know already (obtained the same way above) that:

 

S(n)^1 = n(n+1)/2

 

S(n)^2 = n(n+1)(2n+1)/6

 

S(n)^3 = n^2*(n+1)^2/4

 

The final result is:

 

S(n)^4 = n(6n^4 +15n^3 +10n^2 - 1)/30

 

Edited on December 28, 2008, 7:15 pm

Edited on December 28, 2008, 7:18 pm

Posted by pcbouhid on 2008-12-28 19:13:55