Series problem

[Sweatyapples]

With the patterns noticed in:
1^2 + 2^2 + 3^2 + .. n^2
1^3 + 2^3 + 3^3 + .. + n^3
1^4 + 2^4 + 3^4 + .. n^4

formulate a conjecture for the series, 1^k + 2^k + 3^k + 4^k +.. + n^k
This is what I have so far.

sum r^2 = (n^2/2) + (n/2)
sum r^3 = (n^3/3) + (n^2/2) + (n/6)
sum r^4 = (n^5/5) + (n^4/2) + (n^3/3) - (n/30)

Therefore sum n^k = (1/(k+1))(n^(k+1)) + 1/2(n^k) + .. +
I'm quite confused regarding anything after that. Any help would be appreciated.

 

[Steven G]

With the patterns noticed in:
1^2 + 2^2 + 3^2 + .. n^2
1^3 + 2^3 + 3^3 + .. + n^3
1^4 + 2^4 + 3^4 + .. n^4

formulate a conjecture for the series, 1^k + 2^k + 3^k + 4^k +.. + n^k
This is what I have so far.

sum r^2 = (n^2/2) + (n/2)
sum r^3 = (n^3/3) + (n^2/2) + (n/6)
sum r^4 = (n^5/5) + (n^4/2) + (n^3/3) - (n/30)

Therefore sum n^k = (1/(k+1))(n^(k+1)) + 1/2(n^k) + .. +
I'm quite confused regarding anything after that. Any help would be appreciated.

First you have all formulas wrong!!
sum r^1 = (n^2)/2 + (n/2)
sum r^2 = (n^3)/3 + (n^2)/2 + (n/6)
sum r^3 = (n^4)/4 + (n^3)/2 + (n^2)/4

 

[Sweatyapples]

First you have all formulas wrong!!
sum r^1 = (n^2/2) + (n/2)
sum r^2 = (n^3/3) + (n^2/2) + (n/6)
sum r^3 = (n^5/5) + (n^4/2) + (n^3/3) - (n/30)

My bad,
sum r^1 = (n^2/2) + (n/2)
sum r^2= (n^3/3) + (n^2/2) + (n/6)
sum r^3 = (n^4/4) + (n^3/2) + (n^2/4).
sum r^4 = (n^5/5) + (n^4/2) + (n^3/3) - (n/30)

 

[HallsofIvy]

My bad,
sum r^1 = (n^2/2) + (n/2)
sum r^2= (n^3/3) + (n^2/2) + (n/6)
sum r^3 = (n^4/4) + (n^3/2) + (n^2/4).
sum r^4 = (n^5/5) + (n^4/2) + (n^3/3) - (n/30)

So surely you see that sum r^k would have a first term of n^(k+ 1)/(k+ 1) and a second term of n^k/2. Now, what about that last term?

 

[Sweatyapples]

So surely you see that sum r^k would have a first term of n^(k+ 1)/(k+ 1) and a second term of n^k/2. Now, what about that last term?

I'm aware that the coefficients of the last term are bernoulli numbers, however I dont quite understand the (p+1, k) part of the equation,