The formula is too complicated to remember. It can be remembered much more easily by Implementing Faulhaber's formula. It says,

1ˣ +2ˣ +3ˣ +4ˣ ......nˣ =[1/(1+x)](aB₀nˣ⁺¹+bB₁nˣ+cB₂nˣ⁻¹............yBₓn), where Bₙ is the nth Bernoulli no. and a,b,c,.....y are the consecutive terms of (x+1)th row of Pascal's Triangle.

For e.g.

1¹¹+2¹¹+3¹¹+4¹¹.....7¹¹=(1/12)(1B₀n¹²+12B₁n¹¹+66B₂n¹⁰+220B₃n⁹+495B₄n⁸+792B₅n⁷+924B₆n⁶+792B₇n⁵+495B₈n⁴+220B₉n³+66B₁₀n²+12B₁₁n)

(1+B)ⁿ⁺¹-Bₙ₊₁=0

For e.g.

for,B₁ n=1 so

(1+B)²-B₂=0

⇒1+B₂+2B-B₂=0

⇒1=2B=0

⇒B=-0.5

However in this ( the formula for sums of powers )formula B₂=+0.5]]>

Knowing that:

(excuse me the under subscript is 0)

We can rewrite the series so:

Now i notice that delta_n appears m-n times in the series:

We must work a little more for our formula:

QED]]>

I would say go ahead and give it a shot. Where do you intend to publish?

]]>That does indeed work for what I have tested but there is already a simple closed form for the first sum.

Does your solution have something special that makes it better for something? If so, what?

]]>I haven't been here for a while, but now i'm back with something new. I found a formula that give the result of the partial sum of the series:

For each k positive integer. With recursion i mean: do you want the partial sum formula for n=3? You need to know the partial sum formula for n=2 and for that you need partial sum formula for n=1;etc.

This is the formula:

It works perfectly!

Before I publish the proof i really would like your judge:is it a useful formula? Or it's less interesting than i think?

I thank you for every answer.