Thread: [SOLVED] Faulhaber's formula and series

In my assignment, I am asked to find the general formulas for the following series:

1^2+2^2+3^2+4^2+......+n^2
1^3+2^3+3^3+4^3+......+n^3
1^4+2^4+3^4+4^4+......+n^4

Having worked out the general formulas for the above, i am supposed to note patterns and hence formulate a conjecture for the series

1^k+2^k+3^k+4^k+......+n^k


Having worked out the general formulas:


1^2+2^2+3^2+4^2+......+n^2 = 1/6 n*(n+1)*(2n+1)
1^3+2^3+3^3+4^3+......+n^3 = 1/4 [n*(n+1)]^2
1^4+2^4+3^4+4^4+......+n^4 = 1/30 n(n+1)(2n+1)(3n^2+3n-1)

i have also done some research and found out that the answer to my question is FAULHABER's FORMULA.. but unfortunately I need to be able to note observations and show how to arrive at faulhaber's formula.

Thanks in advance!!

Originally Posted by heather123

In my assignment, I am asked to find the general formulas for the following series:

1^2+2^2+3^2+4^2+......+n^2
1^3+2^3+3^3+4^3+......+n^3
1^4+2^4+3^4+4^4+......+n^4

Having worked out the general formulas for the above, i am supposed to note patterns and hence formulate a conjecture for the series

1^k+2^k+3^k+4^k+......+n^k


Having worked out the general formulas:


1^2+2^2+3^2+4^2+......+n^2 = 1/6 n*(n+1)*(2n+1)
1^3+2^3+3^3+4^3+......+n^3 = 1/4 [n*(n+1)]^2
1^4+2^4+3^4+4^4+......+n^4 = 1/30 n(n+1)(2n+1)(3n^2+3n-1)

i have also done some research and found out that the answer to my question is FAULHABER's FORMULA.. but unfortunately I need to be able to note observations and show how to arrive at faulhaber's formula.

Thanks in advance!!

Try the following conjecture:

S(n) = 1^k+2^k+3^k+4^k+......+n^k

is a ploynomial of degree k+1 in n.

RonL